When the switch is opened, the conducting link between the cell and the bulb is broken. This means the electric circuit is no longer closed (continuous). As a result, the electric charges (electrons) in the metallic wire stop flowing — there is no electric current in the circuit. Since no current flows through the bulb's filament, no heat is produced in it, so the filament does not get hot enough to emit light and the bulb stops glowing.

Source: Chapter 11 – Electricity, Section 11.1 Electric Current and Circuit; Section 11.7.1 Practical Applications of Heating Effect

---

Examiners look for three distinct points (1 mark each):

1. Opening the switch breaks the closed/continuous path (circuit is incomplete).
2. Charges/electrons stop flowing → no electric current.
3. No current means no heating of filament → bulb does not glow.

Avoid vague phrases like "electricity stops." Be specific: it is the flow of charges that stops, and it is the heating effect of current that makes the filament glow. Linking these two ideas explicitly secures full marks.

When electricity was first studied, electrons were unknown. Scientists assumed current was due to the flow of positive charges, so the direction of current was defined as the direction of positive charge flow — from the positive terminal to the negative terminal of a cell. This convention was established before electrons were discovered.

Practical consequence: In circuit analysis, conventional current is always taken to flow from the positive to the negative terminal (through the external circuit), opposite to the actual electron flow. This does not affect calculations, as the results remain the same regardless of which direction is chosen as positive.

Source: Chapter 11, Section 11.1 — Electric Current and Circuit

---

• Key fact examiners look for: The convention was set historically, before electrons were known — state this clearly.
• Mention that current direction = opposite to electron flow.
• The practical consequence: circuits are analysed with conventional current (+ to −), and all laws (Ohm's law, Kirchhoff's laws) work correctly with this convention.
• Don't write a long essay — two short paragraphs covering "why" and "consequence" is sufficient for 3 marks.

(i) I = Q/t = 180 C / (2 × 60) s = 180/120 = 1.5 A

(ii) Using Ohm's Law: R = V/I = 12/1.5 = 8 Ω

Ohm's Law states that the potential difference across a conductor is directly proportional to the current through it at constant temperature (V = IR).

• Convert time to seconds: 2 min = 120 s.
• Formula for current: I = Q/t.
• For resistance, apply Ohm's Law: R = V/I.
• Always name the law explicitly — examiners award a separate mark for it.

(i) The ammeter has nearly zero resistance. When connected in parallel with the 10 Ω resistor, the parallel combination has an equivalent resistance ≈ 0 Ω.

Current through the ammeter branch:

$$I = \frac{V}{R} = \frac{6 \text{ V}}{0 \text{ Ω}} \approx \text{very large (extremely high) current}$$

Almost the entire current flows through the ammeter branch (short circuit).

(ii) This causes a short circuit — the circuit resistance drops nearly to zero, so an extremely large current flows from the cell. This can damage the ammeter, the cell, or cause overheating.

This connection is incorrect because an ammeter must be connected in series to measure current. In parallel, it bypasses the resistor (its near-zero resistance provides an easier path), so it neither measures the correct current through the resistor nor protects the circuit.

Source: Chapter 11, Section 11.6.2 – Resistors in Parallel

---

• Examiners expect you to apply Ohm's law (I = V/R) with R ≈ 0 for the ammeter, giving a dangerously large current — the key concept is short circuit.
• You must state the correct usage (ammeter in series) to explain why parallel connection is wrong.
• Part (ii) carries the reasoning marks — mention both the effect (excessive current/short circuit) and the reason (near-zero resistance of ammeter bypasses the resistor).

(a) The chemical action within the cell generates a potential difference across its terminals. When connected to a resistor, this potential difference sets the charges in motion through the conductor, producing an electric current. It acts like an "electric pressure" that drives electrons around the circuit.

(b) Inside the cell, chemical energy is converted into electrical energy to sustain the current. When internal resistance is considered, some potential difference is used up in driving current through the cell itself (voltage drop = $I \times r$). Thus, the terminal voltage available across the external resistor becomes $V = \varepsilon - Ir$, which is less than the cell's EMF.

Source: Chapter 11, Sections 11.2 and 11.7

---

• For part (a), the key phrase from the textbook is: "the potential difference sets the charges in motion in the conductor." Use the water-pressure analogy if needed but keep it brief.
• For part (b), examiners expect two things: the energy transformation (chemical → electrical) AND the reason for voltage drop (internal resistance consumes part of the EMF). Writing $V = \varepsilon - Ir$ earns the formula mark even if not explicitly asked.
• Avoid writing long paragraphs; concise, pointed sentences score better in board exams.

(a)(i) Using Ohm's law: $R = V/I = 9\text{ V} / 3\text{ A} = \mathbf{3\ \Omega}$

(a)(ii) Work done = Potential difference × Charge
$$W = V \times Q = 9\text{ V} \times 6\text{ C} = \mathbf{54\ J}$$

(b) The student's claim is incorrect. Potential difference across a resistor is not a fixed property of the resistor alone. By Ohm's law, $V = IR$ — it depends on both the resistance and the current through it. The current itself depends on the total resistance of the circuit and the EMF of the source. For example, adding resistors in series reduces the current, which reduces the potential difference across any individual resistor, even though its resistance stays the same.

Source: Chapter 11, Section 11.4 (Ohm's Law) and Section 11.6.1 (Resistors in Series)

---

• Part (a)(i): Straightforward application of $R = V/I$. Always write the formula first.
• Part (a)(ii): Work done $W = VQ$ (since $V = W/Q$ by definition). Don't confuse this with $W = VIt$; here charge $Q$ is directly given.
• Part (b): Examiners want you to clearly say the claim is wrong, state that $V = IR$ means V depends on current, and link current to circuit conditions. One concrete example (series circuit changing current) earns full credit. Avoid vague answers like "it depends on other things."

Ohm's Law: The potential difference across the ends of a resistor is directly proportional to the current through it, provided its temperature remains constant.

Since this is a 1-mark question, state the law in one line and mention the one key condition (constant temperature). Examiners expect both parts — the law itself and the condition — to award full marks.

From the straight-line V–I graph passing through the origin, two conclusions are:

1. Ohm's law is obeyed: The potential difference (V) across the nichrome wire is directly proportional to the current (I) through it (V ∝ I), as the ratio V/I remains constant.

1. The resistance is constant: Since V/I = R = constant, the resistance of the nichrome wire does not change with varying current or voltage (at constant temperature).

Source: Chapter 11, Section 11.4 – Ohm's Law

Examiners expect two distinct, clearly stated points. The key ideas are: (i) direct proportionality → Ohm's law is followed, and (ii) the constant slope of the line means resistance is fixed. Avoid vague statements like "it's a good conductor" — stick to what the graph directly tells you.

Step 1: Find resistance using Ohm's law (V = IR)

$$R = \frac{V}{I} = \frac{4\text{ V}}{0.5\text{ A}} = 8\ \Omega$$

Step 2: New potential difference = 2 × 4 = 8 V

Since resistance remains constant (same resistor):

$$I_{new} = \frac{V_{new}}{R} = \frac{8\text{ V}}{8\ \Omega} = \mathbf{1\ A}$$

Justification: By Ohm's law, $V \propto I$ (R constant). Doubling the potential difference doubles the current.

Source: Chapter 11, Section 11.4 Ohm's Law

---

• Examiners expect two clear steps: (1) calculate R from initial values, (2) apply Ohm's law with new V.
• Always state the justification explicitly — "V ∝ I when R is constant" is the key Ohm's law statement that earns the second mark.
• Don't just give the answer; show the formula and substitution for full credit.

The condition is that the temperature of the metallic wire must remain constant.

This is necessary because resistance depends on temperature. If the temperature rises (due to heating by current), the resistance of the wire changes. Ohm's law states V ∝ I only when resistance R remains constant, which requires constant temperature.

Source: Chapter 11, Section 11.4 – Ohm's Law

---

The examiner expects two things for 2 marks:

1. The condition – constant/same temperature (1 mark)
2. Why it is necessary – because resistance changes with temperature; V ∝ I holds only if R is constant (1 mark)

The textbook phrasing is: "provided its temperature remains the same" — use this or paraphrase it closely. Don't just state the law; you must explain the reason to earn full marks.

A rheostat is a variable resistor. It works by changing the effective length of the resistance wire in the circuit — a longer length increases resistance, a shorter length decreases it. Since resistance depends directly on the length of the conductor, sliding the contact alters the resistance without touching the source.

By Ohm's law: $I = \dfrac{V}{R}$

The source voltage $V$ remains constant. When the rheostat increases resistance $R$, current $I$ decreases; when it decreases $R$, current increases. Thus, the rheostat controls current by varying the resistance of the circuit.

Source: Chapter 11 (Electricity), Sections 11.3 and 11.4

---

• Examiners expect three clear points: (1) what a rheostat does physically (changes effective length → changes resistance), (2) the relevant property (resistance ∝ length of conductor), and (3) Ohm's law to show how changed resistance controls current at fixed voltage.
• Writing the formula $I = V/R$ is essential for full marks — it directly links resistance to current.
• Don't confuse "voltage of the source" with "potential difference across the rheostat"; the source EMF is unchanged, which is what the question asks about.

Since P and Q are made of the same material, their resistivity (ρ) is the same. Let length = $l$ and area of P = $A$, so area of Q = $2A$.

Using $R = \rho \dfrac{l}{A}$:

$$R_P = \rho \frac{l}{A}, \quad R_Q = \rho \frac{l}{2A} = \frac{1}{2} R_P$$

So $R_Q$ is half the resistance of $R_P$.

Since both are connected to the same battery (same V), by Ohm's law $I = V/R$:

$$\frac{I_Q}{I_P} = \frac{R_P}{R_Q} = 2$$

Conclusion: The current through Q is twice the current through P.

Source: Chapter 11, Section 11.5 – Factors on which the resistance of a conductor depends

---

• The key formula is $R = \rho l/A$. Same material → same ρ; same length → only area differs.
• Q has double the area, so half the resistance.
• Same voltage, half the resistance → double the current (Ohm's law: $I = V/R$).
• Examiners award marks for: writing the formula, substituting correctly for both resistors, stating the ratio of resistances, and applying Ohm's law to get the current ratio. Don't skip steps.

(a) The student's claim is incorrect. According to Ohm's law, $V = IR$, which gives $I = V/R$. A high resistance reduces the current but does not make it zero — current will still flow as long as a potential difference is applied. For example, the filament of an electric bulb has a high resistance (~1000 Ω), yet when connected to 220 V, a current of 0.22 A flows through it. Current becomes zero only if the circuit is broken, not merely because resistance is high.

(b)

• Heater: $I = V/R = 200/50 = \mathbf{4\ A}$
• Bulb: $I = V/R = 200/1000 = \mathbf{0.2\ A}$

The heater draws more current, by a factor of 20.

Since power $P = I^2R = V^2/R$, a lower resistance draws more current and dissipates more heat. Heating appliances (irons, heaters) need large amounts of heat energy, so they are designed with low resistance to allow high current and maximum heat production.

Source: Chapter 11 – Electricity, Sections 11.3 (Ohm's Law) and 11.7.1 (Heating Effect Applications)

---

• Part (a): Examiners want you to explicitly state Ohm's law as $I = V/R$, identify the logical flaw (high R ≠ zero I), and give a concrete numerical or real-life example.
• Part (b): Show both calculations clearly with formula, substitution, and result. State which draws more current and calculate the factor (4 ÷ 0.2 = 20). The final explanation linking $P = V^2/R$ to design logic earns the last conceptual mark.
• Do not write vague statements like "resistance opposes current" without connecting it to Ohm's law mathematically.

Using the formula $R = \rho \dfrac{l}{A}$, the original resistance is:
$$R = \rho \frac{l}{A}$$

When the wire is stretched, the new length = $2l$ and new area = $A/2$ (volume constant).

New resistance:
$$R' = \rho \frac{2l}{A/2} = \rho \frac{2l \times 2}{A} = 4\rho \frac{l}{A} = 4R$$

The new resistance is 4R.

Reasoning: Doubling the length doubles the resistance, and halving the area doubles it again. Both effects multiply, giving $2 \times 2 = 4$ times the original resistance.

Source: Chapter 11, Section 11.5 – Factors on which the resistance of a conductor depends.

---

• Examiners expect you to state the formula $R = \rho \dfrac{l}{A}$, apply it to both old and new dimensions, and show the algebraic steps clearly.
• The key insight is that stretching changes both length and area simultaneously, so both changes must be accounted for — this is what the volume-constant condition tells you.
• Writing the final answer as 4R (in terms of the original) is sufficient; no numerical value is needed.
• Avoid skipping steps — in a 3-mark question, 1 mark is typically for the formula, 1 for correct substitution, and 1 for the final answer.

Given:

• Length of P = 2l, Length of Q = l
• Diameter of P = 2d, Diameter of Q = d → Radius of P = 2r, Radius of Q = r

Using the formula $R = \rho \dfrac{l}{A}$, where $A = \pi r^2$:

$$R_P = \rho \cdot \frac{2l}{\pi (2r)^2} = \rho \cdot \frac{2l}{4\pi r^2} = \frac{\rho l}{2\pi r^2}$$

$$R_Q = \rho \cdot \frac{l}{\pi r^2}$$

$$\frac{R_P}{R_Q} = \frac{\rho l / 2\pi r^2}{\rho l / \pi r^2} = \frac{1}{2}$$

∴ Ratio of resistance of P to Q = 1 : 2

Source: Chapter 11 (Electricity), Section 11.5 – Factors on which the resistance of a conductor depends

---

• The key formula is $R = \rho \dfrac{l}{A}$, where $A = \pi r^2 = \pi(d/2)^2$.
• Doubling the length doubles R, but doubling the diameter quadruples the area, which quarters R. Net effect: $2 \times \frac{1}{4} = \frac{1}{2}$.
• Examiners expect you to substitute values clearly and show the cancellation step. Writing the final ratio in the form 1 : 2 is mandatory.

The resistance of a conductor is given by:

$$R = \rho \frac{l}{A}$$

where $l$ is length, $A$ is area of cross-section, and $\rho$ is the resistivity of the material.

Effect of length: Since $R \propto l$, doubling the length doubles the resistance. A longer wire means electrons must travel farther, encountering more opposition.

Effect of area: Since $R \propto \dfrac{1}{A}$, doubling the cross-sectional area halves the resistance. A wider wire provides more paths for electrons to flow, reducing opposition to current.

Thus, resistance depends directly on length and inversely on area of cross-section.

Source: Chapter 11, Section 11.5 – Factors on which the resistance of a conductor depends.

---

• Examiners expect the formula $R = \rho \frac{l}{A}$ to be written — this is the core of the answer and earns a mark directly.
• One mark each for explaining the effect of length and area using the proportionality relationships ($R \propto l$ and $R \propto 1/A$).
• Avoid lengthy analogies; state the relationship, then give a brief physical reason. That's all that's needed for 3 marks.

(A) Halving its length and doubling its cross-sectional area

Using $R = \rho \dfrac{l}{A}$: new resistance $= \rho \dfrac{l/2}{2A} = \dfrac{1}{4}\rho\dfrac{l}{A} = \dfrac{R}{4}$.

Source: Chapter 11, Section 11.5 (Example 11.6)

---

The formula $R = \rho \frac{l}{A}$ shows resistance is directly proportional to length and inversely proportional to area. Halving l gives factor ½; doubling A gives another factor ½ — together reducing R to ¼. This is directly solved in Example 11.6 of the textbook. For MCQs, always substitute the changes into the formula quickly to verify.

Nichrome is preferred over copper as a heating element due to the following reasons:

1. High resistivity: Nichrome has resistivity $100 \times 10^{-6}$ Ω m, far greater than copper's $1.62 \times 10^{-8}$ Ω m. Higher resistivity means more heat is generated (by Joule's heating, $H = I^2Rt$) for the same current.

1. Does not oxidise at high temperatures: Alloys like nichrome do not burn or oxidise readily at high temperatures, making them suitable for repeated heating.

1. Copper's low resistivity makes it an excellent conductor but a poor heat generator, so it is used for transmission lines, not heating devices.

Source: Chapter 11, Section 11.5 (Table 11.2) and Section 11.7.1

---

Examiners expect two clear points: (i) high resistivity → more heat generated, with values from Table 11.2 as evidence, and (ii) alloys don't oxidise at high temperatures. Mentioning Joule's heating formula adds precision. Avoid writing lengthy paragraphs — crisp, labelled points score better in 3-mark answers.

The claim is incorrect.

Resistance depends not only on length and area of cross-section but also on the nature (material) of the conductor, given by $R = \rho \frac{l}{A}$, where $\rho$ is the electrical resistivity of the material. Different materials have different resistivities (e.g., copper has $1.62 \times 10^{-8}\ \Omega\text{m}$, nichrome has $100 \times 10^{-6}\ \Omega\text{m}$). So changing the material changes the resistance, which in turn changes the current ($I = V/R$).

Source: Chapter 11, Section 11.5 – Factors on which the resistance of a conductor depends

---

• Examiners want you to state clearly that the claim is wrong, then give the reason using the formula $R = \rho l/A$.
• Name resistivity ($\rho$) as the material-dependent factor — this is the key term.
• Quoting one example of contrasting resistivity values (e.g., copper vs. nichrome) strengthens the answer and shows you've read the table.
• Linking back to $I = V/R$ shows you understand why current changes — good for full marks.

Electrical resistivity is the resistance offered by a material of unit length and unit cross-sectional area. It is given by $\rho = \frac{RA}{l}$, with SI unit ohm-metre (Ω m).

Yes, since both wires have identical dimensions (same $l$ and $A$), the wire with higher resistance is the poorer conductor. However, resistance depends on dimensions too. Resistivity is an intrinsic property of the material, independent of shape or size — making it more suitable for comparing materials directly.

Source: Chapter 11, Section 11.5 – Factors on which the resistance of a conductor depends

---

• Define resistivity using the formula $\rho = RA/l$ and state its unit (Ω m) — this alone is worth 1 mark.
• For the second part, acknowledge that yes, since dimensions are identical, resistance comparison works here, but then explain why resistivity is better: it is a material property independent of dimensions. This distinction is what examiners look for in the second mark.
• Don't write lengthy explanations — two focused points are enough for 2 marks.

Wire Y (manganin) offers greater opposition to electron flow.

This is because resistance depends on the material's resistivity ($R = \rho \dfrac{l}{A}$). Since both wires have identical dimensions (same $l$ and $A$), the resistance depends only on resistivity.

Manganin has resistivity approximately 2700 times that of copper. Therefore:

$$\frac{R_Y}{R_X} = \frac{\rho_{\text{manganin}}}{\rho_{\text{copper}}} = 2700$$

The resistance of wire Y is 2700 times greater than that of wire X.

Source: Chapter 11, Section 11.5 – Factors on which the resistance of a conductor depends

---

• The key formula examiners expect is $R = \rho \dfrac{l}{A}$. Since $l$ and $A$ are identical, the ratio of resistances equals the ratio of resistivities directly.
• State which wire clearly first, then justify with the formula — this earns the first mark.
• The factor calculation is straightforward substitution — show the ratio setup to secure the final mark.
• Manganin is an alloy; alloys generally have much higher resistivity than pure metals, which is why they are used in heating devices.

When a nichrome wire is heated, its resistivity increases. Since resistance $R = \rho \dfrac{l}{A}$, an increase in resistivity causes an increase in resistance.

By Ohm's law: $I = \dfrac{V}{R}$

Since the voltage V remains the same but resistance R increases, the current through the wire decreases. Current is inversely proportional to resistance — if resistance increases, current decreases proportionally.

Source: Chapter 11 (Electricity), Sections 11.4 and 11.5

• Key relationships to quote: $R = \rho \frac{l}{A}$ (links resistivity to resistance) and $I = V/R$ (Ohm's law linking current to resistance and voltage).
• Examiners want you to clearly state: resistivity ↑ → resistance ↑ → current ↓ (at same V), with the formulas justifying each step.
• Don't just say "current decreases" — you must justify why using both equations to earn full 3 marks.

Total resistance of wire = 12 Ω. When bent into a square, each side has resistance = 12/4 = 3 Ω.

Between two diagonally opposite corners (say A and C), the wire splits into two paths:

• Path 1: A→B→C = 3 + 3 = 6 Ω
• Path 2: A→D→C = 3 + 3 = 6 Ω

These two 6 Ω resistors are in parallel:

$$\frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6}$$

$$R_p = 3 \text{ Ω}$$

The resistance between diagonally opposite corners = 3 Ω.

Source: Chapter 11, Section 11.6 (Resistance of a System of Resistors)

---

• First find resistance of each side by dividing total resistance by 4.
• Identify the two parallel paths between opposite corners (each path has 2 sides in series = 6 Ω).
• Apply the parallel combination formula. Examiners award marks for: correctly finding each side's resistance, correctly identifying series/parallel arrangement, and the final calculation.

(i) Series combination:

$R_s = 6 + 6 + 6 = 18\ \Omega$

$I_s = V/R_s = 12/18 = 0.67\ \text{A}$

Parallel combination:

$\frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6}$, so $R_p = 2\ \Omega$

$I_p = V/R_p = 12/2 = 6\ \text{A}$

(ii) Ratio of currents:

$$\frac{I_p}{I_s} = \frac{6}{0.67} = 9:1$$

In series, equivalent resistance increases (equals sum of all resistances = 18 Ω). In parallel, equivalent resistance decreases below the smallest individual resistance (= 2 Ω). Since $I = V/R$ and voltage is the same, the 9-fold lower resistance in parallel draws 9 times more current.

(iii) Two reasons parallel connection is preferred in households:

1. Each appliance gets the full supply voltage (220 V), so every device operates at its rated capacity independently.
2. If one appliance fails, the others continue to work, since each is on a separate branch — unlike series, where one failure breaks the entire circuit.

Source: Chapter 11, Sections 11.6.1 and 11.6.2

---

• For part (i), always show the formula, substitution, and unit — examiners award step marks.
• For part (ii), the ratio $I_p : I_s = 9 : 1$ directly follows from $R_s/R_p = 18/2 = 9$; state this link clearly.
• For part (iii), the textbook explicitly lists these two advantages in Section 11.6.2 — use those exact points. Do not invent new ones.
• Keep calculations neat; a missing unit can cost half a mark.

Total resistance $R_s = 4 + 6 + 10 = 20\ \Omega$. Current $I = V/R_s = 20/20 = 1\ \text{A}$. Potential difference across 6 Ω $= IR = 1 \times 6 = \mathbf{6\ V}$.

In a series circuit, the same current flows through all resistors. Find total resistance, then current using Ohm's law, then apply $V = IR$ to the individual resistor. Examiner expects all three steps shown briefly even in a 1-mark answer.

The ammeter will read 0.3 A. In a series circuit, the current remains the same throughout; it does not change with the position of the ammeter in the circuit.

Source: Chapter 11, Section 11.6.1 – Resistors in Series

The key principle tested here is: current is constant at every point in a series circuit. The examiner wants the correct value (0.3 A) AND the justification (series circuit → same current everywhere). Both parts are needed for full credit even in a 1-mark question. Avoid saying "no calculation needed" as your justification — state the physics law instead.

Resistance opposes the flow of electric current through a conductor. In a series combination, resistors are joined end to end, so the same current must pass through each resistor one after another, experiencing opposition from every resistor in its path.

The equivalent resistance is given by:
$$R_s = R_1 + R_2 + R_3$$

Since all individual resistances are positive values, their sum must be greater than any single resistance alone. Each additional resistor adds more opposition to current flow, so $R_s$ is always greater than $R_1$, $R_2$, or $R_3$ individually.

Source: Chapter 11, Section 11.6.1

---

Examiners look for three things here:

1. What resistance means — it opposes/resists current flow (don't just say "it's a property").
2. Why series adds resistances — same current passes through each resistor sequentially, facing opposition at every stage.
3. The formula $R_s = R_1 + R_2 + R_3$ with the logical conclusion that a sum of positive numbers exceeds any one term.

Avoid vague statements like "resistance increases." Tie it directly to the physics: sequential opposition experienced by the same current.

(a) Current and potential difference:

Total resistance: $R_s = 5 + 10 + 15 = 30\ \Omega$

Current: $I = V/R_s = 30/30 = \mathbf{1\ A}$

Potential difference across each resistor (V = IR):

• $V_1 = 1 \times 5 = \mathbf{5\ V}$
• $V_2 = 1 \times 10 = \mathbf{10\ V}$
• $V_3 = 1 \times 15 = \mathbf{15\ V}$

(b) Resistor dissipating most power:

Using $P = I^2R$ (current is same in series):

• $P_1 = 1^2 \times 5 = 5\ \text{W}$
• $P_2 = 1^2 \times 10 = 10\ \text{W}$
• $P_3 = 1^2 \times 15 = \mathbf{15\ W}$

The 15 Ω resistor dissipates the most power. In a series circuit, since current is constant, $P \propto R$, so the largest resistor dissipates maximum power.

(c) Change in power dissipated in 5 Ω resistor:

New circuit: 5 Ω and 10 Ω in series, V = 30 V

New total resistance: $R_s = 5 + 10 = 15\ \Omega$

New current: $I' = 30/15 = 2\ \text{A}$

New power in 5 Ω: $P' = I'^2 \times 5 = 4 \times 5 = \mathbf{20\ W}$

Previously, $P = 1^2 \times 5 = 5\ \text{W}$. The power dissipated in the 5 Ω resistor increases from 5 W to 20 W (4 times) because removing the 15 Ω resistor reduces total resistance, increasing current.

Source: Chapter 11, Sections 11.6.1 and 11.8

---

• Examiners award 1 mark each for correct $R_s$ and $I$ in part (a), and 1 mark for all three V values.
• In part (b), you must quote the formula $P = I^2R$ and state why the largest R wins (same I throughout series circuit).
• In part (c), you must recalculate both current and power — just stating "power increases" without numbers earns no marks. Show the before and after comparison explicitly.

Normally, both cells (each 1.5 V) add up in series to give a total EMF of 3 V, producing a current of $I = \frac{3}{5} = 0.6$ A through the bulb.

When one cell is reversed, it opposes the other. The net EMF = 1.5 V − 1.5 V = 0 V. By Ohm's law, $I = \frac{V}{R} = \frac{0}{5} = 0$ A.

The bulb goes out completely — no current flows.

This is consistent with series-circuit principles: in a series circuit the same current flows throughout, and the total potential difference equals the algebraic sum of all individual EMFs. With equal and opposing cells, they cancel, leaving zero net driving voltage and hence zero current.

Source: Chapter 11 – Electricity, Section 11.6.1 Resistors in Series

---

• The key series principle here is that EMFs in series add algebraically — a reversed cell subtracts.
• Examiners expect the numerical calculation ($I = V/R$) showing zero current, plus the reasoning using series-circuit rules.
• Do not just say "current decreases" — two identical cells reversed give exactly zero net EMF and zero current, not merely a reduced current.

In a series circuit, the same current flows through all components, but the total potential difference (voltage) is shared (divided) among them.

This makes series arrangement unsuitable for domestic appliances because:

1. Different appliances need different voltages to operate correctly. In series, the supply voltage is split across each appliance, so no appliance gets the required voltage.
2. If one appliance is switched off or fails, the circuit breaks, and all other appliances stop working.
3. All appliances must operate at the same current, which is not practical since different appliances (e.g., refrigerator and ceiling fan) have different current requirements.

Thus, a series circuit cannot provide independent control or correct operating conditions for domestic appliances.

Source: Chapter 11 – Electricity, Section 11.6.1 (Resistors in Series)

---

Examiners look for three specific disadvantages clearly linked to the properties of series circuits — shared voltage, same current, and circuit breaking on failure. Mention both the electrical property (what happens in series) and why it's a problem for appliances. Don't just state facts — connect them to the real-life problem. This question often appears as Q18(c) in board exams; the answer should be concise but cover all three points to secure full 3 marks.

$\frac{1}{R_p} = \frac{1}{6}+\frac{1}{12}+\frac{1}{4} = \frac{2+1+3}{12} = \frac{6}{12}$, so $R_p = 2\ \Omega$; total current $I = 12/2 = 6\ \text{A}$. The equivalent resistance (2 Ω) is less than the smallest resistor (4 Ω), as expected in parallel combination.

Source: Chapter 11, Section 11.6.2

---

This is actually a multi-part calculation question compressed into 1 mark, so keep each step brief. Examiners look for: correct parallel formula, correct $R_p = 2\ \Omega$, correct $I = 6\ \text{A}$, and the concluding observation that parallel equivalent resistance is always less than the smallest individual resistor. Show the reciprocal calculation clearly — it's the key formula for parallel circuits.

In a parallel combination, all resistors are connected between the same two points, so the same potential difference (V) acts across each of them.

However, by Ohm's law, current through each resistor is $I = V/R$. Since the resistors have different resistances, the current through each is different:

$$I_1 = \frac{V}{R_1}, \quad I_2 = \frac{V}{R_2}, \quad I_3 = \frac{V}{R_3}$$

A resistor with lower resistance allows more current, and one with higher resistance allows less current, even though the potential difference across each is the same.

Source: Chapter 11, Section 11.6.2 – Resistors in Parallel

---

• Key concept 1: In parallel, both ends of every resistor are connected to the same two nodes → same V across each. This is a direct consequence of the circuit arrangement.
• Key concept 2: Ohm's law ($I = V/R$) then gives different currents because R values differ.
• Examiners award 1 mark for explaining why V is the same and 1 mark for applying Ohm's law to explain why I differs.
• Write the formula $I = V/R$ explicitly — it shows you're grounding the answer in Ohm's law, not just stating it vaguely.

The student is incorrect. In a parallel circuit, the same voltage (20 V) acts across both resistors, but current depends on resistance: higher resistance means lower current (by Ohm's law, $I = V/R$).

Calculation:

$$I_P = \frac{V}{R_P} = \frac{20}{10} = 2 \text{ A}$$

$$I_Q = \frac{V}{R_Q} = \frac{20}{40} = 0.5 \text{ A}$$

Since P has lower resistance (10 Ω), more current flows through P (2 A), not Q (0.5 A). The student confused equal voltage with equal current — while voltage is the same across parallel branches, current is inversely proportional to resistance.

Source: Chapter 11, Section 11.6.2 — Resistors in Parallel

---

• Examiners expect you to clearly state the student is wrong, give the reason (Ohm's law in parallel), and show both calculations.
• Key concept: In parallel, voltage is the same across all branches, but current divides — more current takes the path of least resistance.
• Write $I = V/R$ explicitly; don't skip steps in a calculation question.
• The phrase "inversely proportional to resistance" scores a conceptual mark even without numbers.

In a parallel combination, the same potential difference V acts across each resistor. Each branch provides an additional path for current, so total current increases. Since $R_p = V/I$, a larger total current means a smaller equivalent resistance.

The formula confirms this:
$$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots$$

Since $\frac{1}{R_p}$ is greater than any individual term (e.g., $\frac{1}{R_1}$), it follows that $R_p < R_1$ and $R_p < R_2$.

Numerical Example: Take $R_1 = 6\ \Omega$, $R_2 = 3\ \Omega$ (smallest = 3 Ω).

$$\frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$$

$$R_p = 2\ \Omega$$

$R_p = 2\ \Omega < 3\ \Omega$ (smallest resistor). ✓

Source: Chapter 11, Section 11.6.2 — Resistors in Parallel

---

• Key idea: Parallel connection adds extra current paths without changing voltage, so total current rises → equivalent resistance falls. This is the physical reason, and examiners expect it.
• Formula logic: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$ means $\frac{1}{R_p} > \frac{1}{R_{\min}}$, so $R_p < R_{\min}$. State this explicitly.
• Always verify with numbers when the question asks for it — full marks require both reasoning and the example.
• Keep the calculation neat and boxed/highlighted so the examiner spots it quickly.

Step 1: Find total current drawn by the three appliances (in parallel, V = 220 V)

$$I_1 = \frac{220}{100} = 2.2 \text{ A}, \quad I_2 = \frac{220}{50} = 4.4 \text{ A}, \quad I_3 = \frac{220}{500} = 0.44 \text{ A}$$

$$I_{total} = 2.2 + 4.4 + 0.44 = 7.04 \text{ A}$$

Step 2: The electric iron draws the same current = 7.04 A

Step 3: Resistance of electric iron

$$R_{iron} = \frac{V}{I} = \frac{220}{7.04} \approx 31.25 \text{ Ω}$$

The resistance of the electric iron is approximately 31.25 Ω.

Source: Chapter 11 (Electricity), Section 11.6 – Resistance of a System of Resistors

---

• In a parallel circuit, each appliance has the full 220 V across it, so use $I = V/R$ separately for each.
• Add all three currents to get the total; the iron draws this same total current.
• Apply Ohm's law ($R = V/I$) to find the iron's resistance.
• Examiners award marks for: correct individual currents (1 mark), correct total current (1 mark), and correct resistance of iron (1 mark). Show each step clearly.

Household appliances are wired in parallel for the following reasons:

1. Each appliance gets the same voltage (220 V): In a parallel circuit, every appliance is connected directly across the supply, so it operates at its rated voltage. In series, the supply voltage would be divided, causing appliances to work below their rated capacity.

1. Independent operation: Each appliance has its own separate branch. It can be switched on or off independently without affecting the others. In series, if one appliance fails or is switched off, the circuit breaks and all appliances stop working.

Source: Electricity, Chapter 11 (Exercise Q.18c; Section 11.6)

---

Examiners expect two clearly distinct points. The two most scoring reasons are (i) equal/full voltage across each device and (ii) independent operation. Avoid repeating the same idea in different words — that scores only 1 mark. Mention the contrast with series briefly to strengthen each point. "Same rated voltage" and "independent switching/failure" are the two standard answers in the NCERT context.

Given: EMF (E) = 24 V, internal resistance (r) = 2 Ω, R₁ = 10 Ω, R₂ = 15 Ω, R₃ = 30 Ω (in parallel)

(a) Equivalent resistance of parallel combination:

$$\frac{1}{R_p} = \frac{1}{10} + \frac{1}{15} + \frac{1}{30} = \frac{3+2+1}{30} = \frac{6}{30} = \frac{1}{5}$$

$$\therefore R_p = 5 \text{ Ω}$$

(b) Total current from battery:

Total resistance = $R_p + r = 5 + 2 = 7$ Ω

$$I = \frac{E}{R_p + r} = \frac{24}{7} \approx 3.43 \text{ A}$$

(c) Potential difference across parallel combination:

Voltage drop across internal resistance = $I \times r = 3.43 \times 2 = 6.86$ V

$$V_{parallel} = E - Ir = 24 - 6.86 \approx 17.14 \text{ V}$$

(d) Current through 30 Ω resistor:

Same potential difference (17.14 V) appears across each branch.

$$I_3 = \frac{V_{parallel}}{R_3} = \frac{17.14}{30} \approx 0.57 \text{ A}$$

Source: Chapter 11, sections 11.6.1 and 11.6.2

---

• Examiners award 1 mark each for parts (a), (b), (c), and (d), with 1 mark for correct formula/method shown at any step.
• Show the formula before substituting values — this earns method marks even if arithmetic slips.
• Key concept: the same PD acts across every branch of a parallel combination, so Ohm's law applies directly to each branch for part (d).
• Internal resistance acts as a series resistor; voltage is "lost" across it, so the terminal voltage (PD across the external circuit) is less than the EMF.

According to Joule's law of heating, heat produced is given by $H = I^2Rt$.

Since both carry the same current, the heat produced depends on resistance. The heating element has very high resistance, so it produces much more heat and glows red-hot. The connecting cord has very low resistance, so negligible heat is produced and it stays cool.

Source: Chapter 11, Section 11.7 – Heating Effect of Electric Current

---

• The key law to quote is Joule's law: $H = I^2Rt$.
• Since current $I$ is the same, the only variable is resistance $R$.
• Examiners expect you to explicitly state that the heating element has high resistance and the cord has low resistance — this is the direct cause.
• Don't write a long explanation; two crisp points earn full marks.

A fuse wire is made of a metal or alloy with a low melting point (e.g., aluminium, lead, or their alloys). During normal operation, the current is within the rated value, so the heat produced (H = I²Rt) is not enough to melt the wire.

When an excessive current flows, heat produced increases rapidly (proportional to I²), raising the fuse wire's temperature above its low melting point. The wire melts quickly, breaking the circuit and protecting the appliance.

The fuse is connected in series so that the entire circuit current passes through it. The material's low melting point ensures fast response to overload, while its rated thickness ensures it survives normal current without melting.

Source: Chapter 11, Section 11.7.1

---

• Examiners expect three distinct points: (1) material has low melting point, (2) normal current → insufficient heat → wire intact, (3) excess current → heat ∝ I² rises sharply → wire melts and breaks circuit.
• Mentioning Joule's law (H = I²Rt) scores a mark and shows understanding.
• "Series connection" is a useful design point but don't spend too many words on it — focus on material + heating effect.
• Avoid vague phrases like "it melts easily" — always link to low melting point and Joule's heating for full marks.

Series Combination:

Total resistance = 2 + 8 = 10 Ω. Let battery voltage = V.

Current through both resistors: $I = V/10$

Heat produced per second (Power) = $I^2R$

• In P: $H_P = I^2 \times 2 = \frac{V^2}{100} \times 2 = \frac{2V^2}{100}$
• In Q: $H_Q = I^2 \times 8 = \frac{V^2}{100} \times 8 = \frac{8V^2}{100}$

Since Q has a higher resistance and current is the same in series, Q produces more heat per second. The student is correct.

Ratio $H_P : H_Q = 2 : 8 = \mathbf{1 : 4}$

---

Parallel Combination:

Both resistors have the same potential difference V across them.

Heat produced per second = $V^2/R$

• In P: $H_P = V^2/2$
• In Q: $H_Q = V^2/8$

Ratio $H_P : H_Q = \frac{V^2}{2} : \frac{V^2}{8} = 4 : 1$

In parallel, P produces more heat per second than Q. The ratio reverses to 4 : 1.

Source: Chapter 11 – Electricity, Section 11.8 Electric Power

---

• In series, current is the same → use $P = I^2R$; the larger resistor dissipates more power.
• In parallel, voltage is the same → use $P = V^2/R$; the smaller resistor dissipates more power.
• Examiners want the formula stated, a numerical ratio calculated, and a clear conclusion for each case. Both parts are mandatory for full marks.

Tungsten for bulb filaments: Tungsten has a very high melting point (3380°C). When current flows, it generates a large amount of heat (Joule's heating). This heat raises the filament to very high temperatures, causing it to emit light. Since tungsten does not melt at such temperatures, it is ideal. However, this same property — converting most electrical energy into heat — makes it unsuitable for transmission lines, where energy loss is undesirable.

Copper/Aluminium for transmission lines: Copper and aluminium have very low resistivity, so they produce minimal heat (low energy loss) when current passes through them. This ensures efficient transfer of electrical energy over long distances with negligible wastage.

Source: Chapter 11, Section 11.7.1 (Practical Applications of Heating Effect of Electric Current)

---

• The key contrast is high resistivity + high melting point (tungsten) vs low resistivity (copper/aluminium).
• For a filament: high resistance → more heat → light emission; high melting point → doesn't melt.
• For transmission: low resistance → less heat loss → efficient energy transfer.
• Examiners expect you to use the term Joule's heating and mention melting point for tungsten and low resistivity/resistance for copper/aluminium. These are the scoring keywords.

Calculation:

Power of electric iron, P = 1.5 kW = 1500 W
Voltage, V = 220 V

Current drawn by the iron:
$$I = \frac{P}{V} = \frac{1500}{220} \approx 6.82 \text{ A}$$

Conclusion: No, the 5 A fuse will not protect the electric iron safely. Since the iron draws approximately 6.82 A, which is greater than the fuse rating of 5 A, the fuse will blow (melt) and break the circuit every time the iron is used. A fuse of higher rating (e.g., 7 A or 10 A) should be used instead.

Source: Chapter 11, Section 11.7.1

---

• The key formula is $I = P/V$. Convert kW to W first (1.5 kW = 1500 W).
• Since the calculated current (6.82 A) exceeds the fuse rating (5 A), the fuse will blow during normal operation — so it does not protect the iron safely; it simply keeps tripping.
• Examiners expect: the formula, correct substitution, correct answer (~6.82 A), and a clear justified conclusion. All three steps carry marks.

(i) Why tungsten is used for a bulb filament (not a heating coil):

In an electric bulb, the filament must become extremely hot so that it emits light. Tungsten has a very high melting point (3380°C), so it does not melt even at the high temperatures needed to produce light. It is also a strong metal that can retain the heat generated. However, tungsten is not preferred for heating coils because it would oxidise easily at high temperatures when exposed to air, making it less suitable for open heating elements.

(ii) Why an alloy is preferred over a pure metal for heating elements:

Alloys such as nichrome are preferred for heating coils because:

• They have higher resistivity than pure metals, so they produce more heat for the same current.
• They have a high melting point, so they do not melt or deform easily at high operating temperatures.
• They do not oxidise (corrode) readily at high temperatures, giving them a longer working life.

Source: Chapter 11, Section 11.7.1 (Practical Applications of Heating Effect of Electric Current)

---

• The textbook (Exercise 18a & 18b) directly asks these two questions, so examiners expect answers rooted in the three properties: melting point, resistivity, and oxidation resistance.
• For (i), the key contrast is: tungsten's high melting point suits the extreme temperatures needed for light emission in a sealed bulb; but heating coils need a material that also resists oxidation in open air — which tungsten doesn't do as well as nichrome.
• For (ii), always mention all three alloy advantages (high resistivity, high melting point, resistance to oxidation) — each is worth marks separately.
• Avoid writing vague statements like "alloys are better" without specifying why.

Step 1: Find the resistance of the kettle.

Using $P = \dfrac{V^2}{R}$:

$$R = \frac{V^2}{P} = \frac{(240)^2}{2000} = \frac{57600}{2000} = 28.8 \ \Omega$$

Step 2: Find power at 120 V.

$$P' = \frac{V'^2}{R} = \frac{(120)^2}{28.8} = \frac{14400}{28.8} = 500 \ \text{W}$$

Is the student correct?
No. The student is wrong. Since $P = \dfrac{V^2}{R}$, power is proportional to the square of voltage. Halving the voltage reduces the power to one-fourth (not one-half): $2000 \times \frac{1}{4} = 500$ W.

Source: Chapter 11 – Electricity, Section 11.8 Electric Power

---

• Examiners expect you to use $P = V^2/R$ (not $P = VI$) when resistance is constant, because R doesn't change but I does.
• The common error the student makes is assuming $P \propto V$, but the correct relation is $P \propto V^2$. Always state this clearly to earn the justification mark.
• Show all three steps: find R, find new P, then answer the conceptual part. Each step carries roughly 1 mark.

Bulbs glow brighter in parallel arrangement.

Equivalent Resistance: In series, the equivalent resistance $R_s = R_1 + R_2 + R_3 = 3R$, which is higher than a single bulb's resistance. In parallel, $\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R}$, giving $R_p = R/3$, which is much lower.

Current Distribution: Since the same battery is used, the lower equivalent resistance in parallel allows a greater total current to flow. Each bulb in parallel receives the full battery voltage, drawing more current individually. In series, the same small current flows through all bulbs and the voltage is divided among them.

Therefore, each bulb receives more power ($P = I^2R$ or $P = V^2/R$) in the parallel arrangement and glows brighter.

Source: Electricity, sections 11.6.1 and 11.6.2

---

• Examiners award marks for three clear points: (1) comparison of equivalent resistance, (2) current distribution/behavior, and (3) conclusion about brightness.
• Always mention both series formula ($R_s = R_1+R_2+R_3$) and parallel formula to show higher vs. lower resistance — this directly earns marks.
• The key insight: in parallel, each bulb gets the full terminal voltage; in series, voltage is shared, so current and power per bulb are less.
• Quoting $P = V^2/R$ or $P = I^2R$ as justification for brightness is a valued finishing touch.

(A) Both A and R are true and R is the correct explanation of A.

In parallel combination, each appliance gets the full supply voltage and operates independently; equivalent resistance decreases, allowing each device to draw its required current.

The textbook (section 11.6.2) explicitly states that in a parallel circuit, the potential difference across each gadget is the same (full supply voltage), each draws its own required current independently, and the total equivalent resistance decreases. Both the Assertion and the Reason are correct, and R directly explains why parallel connection is preferred — making option (A) the right choice.

(i) The physical property responsible is electrical resistivity. Nichrome has a very high resistivity ($100 \times 10^{-6}$ Ω m), whereas copper has a very low resistivity ($1.62 \times 10^{-8}$ Ω m). Since both wires have the same length and cross-sectional area, nichrome has a much higher resistance than copper.

(ii) The power (heat per second) dissipated in a resistor is given by:
$$P = I^2R$$
Since both P and Q carry the same current $I$, the heat generated per second depends directly on resistance $R$. As nichrome has far greater resistance than copper, resistor P dissipates much more heat per second than Q.

Source: Chapter 11 — Sections 11.5 (Factors affecting resistance/resistivity) and 11.7 (Heating effect of electric current)

---

• Part (i) tests knowledge of resistivity as a material property. Examiners expect you to name "resistivity," state that nichrome's resistivity is much higher than copper's, and link this to higher resistance (same l and A).
• Part (ii) requires the formula $P = I^2R$ (from Joule's law). Since I is the same in series, more R → more heat/second. Always write the formula and then reason from it — that's what examiners award marks for.
• Do not confuse resistance (depends on dimensions too) with resistivity (material property only). Part (i) specifically asks for the material property, so "resistivity" is the key term.

(i) Heating element vs. copper wires:

From $R = \rho l/A$, resistance depends on resistivity ($\rho$), length, and cross-sectional area. The heating element of a toaster is made of an alloy (e.g., nichrome) which has very high resistivity ($\sim 100 \times 10^{-6}$ Ω m), whereas copper has very low resistivity ($1.62 \times 10^{-8}$ Ω m). Since $P = I^2R$, for the same current $I$, power dissipated is directly proportional to resistance. The heating element therefore has much higher resistance and dissipates far more heat, becoming very hot, while copper wires have negligible resistance and remain cool.

(ii) Fuse wire — property and design:

The fuse wire is made of a metal or alloy (e.g., aluminium, lead) with a low melting point. During normal operation, the current is within the rated value, so heat produced is insufficient to melt it. During a fault, excess current increases heat generation ($H = I^2Rt$) rapidly, raising the fuse wire's temperature above its melting point — it melts and breaks the circuit, protecting the appliances.

(iii) Why copper is preferred for connecting wires:

Copper has very low resistivity ($1.62 \times 10^{-8}$ Ω m), so its resistance is extremely small. This means minimal power is wasted as heat ($P = I^2R$) in the connecting wires, ensuring efficient transmission of electrical energy to the appliance.

Source: Chapter 11, Sections 11.5 and 11.7.1

---

• Examiners look for explicit linking of $R = \rho l/A$ → high $\rho$ → high $R$ → high $P = I^2R$ for part (i). Just saying "resistance is high" without linking to resistivity loses marks.
• For part (ii), the key phrases are low melting point and the contrast between normal vs. fault current — both must appear.
• For part (iii), keep it brief: low resistivity → low resistance → less heat wasted. Don't repeat part (i) at length.
• Alloys are used for heating elements (not pure metals) because alloys have higher resistivity and don't oxidise at high temperatures — useful to mention if the question asks, but here focus on the contrast with copper.

(i) By Ohm's law, $I = V/R$. If V is tripled and R is unchanged, the current also triples to 3I.

(ii) Resistance $R = \rho L/A$. New length = 2L, new area = A/2.

$$R_{\text{new}} = \rho \cdot \frac{2L}{A/2} = \frac{4\rho L}{A} = 4R$$

The resistance increases by a factor of 4.

(iii) Original current: $I = V/R$

New current: $I_{\text{new}} = \dfrac{3V}{4R} = \dfrac{3}{4} \cdot \dfrac{V}{R} = \dfrac{3I}{4}$

$$\frac{I_{\text{new}}}{I} = \boxed{\frac{3}{4}}$$

Source: Chapter 11, Sections 11.4 and 11.5 (Ohm's Law; Resistance of a Conductor)

---

• Part (i): Direct application of $I = V/R$ — if R is constant, I is proportional to V.
• Part (ii): Examiners expect the formula $R = \rho L/A$ with substitution shown step by step. Both changes (doubled L and halved A) each double the resistance, giving a combined factor of 4.
• Part (iii): Combine both changes: voltage ×3, resistance ×4. The net effect is $3/4$, so the new current is less than the original despite the higher voltage. Always show the ratio explicitly.

(i) Power dissipated in the resistor:
$$P = \frac{V^2}{R}$$

(ii) When R is halved (R becomes R/2):
$$P' = \frac{V^2}{R/2} = \frac{2V^2}{R} = 2P$$

The power dissipated doubles.

(iii) Higher power means a larger current flows through the circuit (since $I = V/R$, halving R doubles the current). A fuse melts when current exceeds its rated value. The doubled current produces excess heat in the fuse wire, causing it to melt and break the circuit.

Source: Chapter 11, Section 11.7 Heating Effect of Electric Current

---

• Examiners award 1 mark each for the three parts. Part (i) needs the correct formula ($V^2/R$); part (ii) needs to show working and state "doubles"; part (iii) needs to link increased current → excess heating → fuse melts.
• The key formula here is $P = V^2/R$ (not $I^2R$), because V is constant in this circuit.
• Always state the conclusion explicitly ("power doubles", "fuse melts") — don't leave the examiner to infer it.

(i) Resistance of each device:

Using $R = V^2/P$:

• Electric iron: $R_1 = \dfrac{(220)^2}{1000} = \dfrac{48400}{1000} = **48.4 \text{ Ω}**$

• Electric bulb: $R_2 = \dfrac{(220)^2}{100} = \dfrac{48400}{100} = **484 \text{ Ω}**$

(ii) Total current in parallel:

Current through iron: $I_1 = \dfrac{220}{48.4} \approx 4.55 \text{ A}$

Current through bulb: $I_2 = \dfrac{220}{484} \approx 0.45 \text{ A}$

Total current $I = I_1 + I_2 = 4.55 + 0.45 = **5 \text{ A}**$

(iii) Series connection:

Total resistance $R_s = 48.4 + 484 = 532.4 \text{ Ω}$

Current $I = \dfrac{220}{532.4} \approx **0.41 \text{ A}**$

Why series is unsuitable:

1. Both devices share the same small current (≈0.41 A), but the iron needs ~4.55 A and the bulb ~0.45 A to work properly — neither device operates at its rated power.
2. If one device fails/fuses, the circuit breaks and the other stops working too.

Source: Chapter 11 — Electricity, Sections 11.6.1 and 11.6.2

---

• The formula $R = V^2/P$ is the quickest route here since rated voltage and power are given directly.
• For part (ii), you can also use $I = P/V$ for each device — same result, equally accepted.
• In part (iii), the series current calculation must come first, then the two distinct reasons. Examiners award marks separately for the calculation and each reason, so state both reasons clearly. The textbook explicitly states that different gadgets need different currents and that series circuits fail completely when one component breaks — use these points.
• Do not confuse "unsuitable because voltage is divided" (that is a consequence) with the key reasons above; stick to what the textbook states.