A student connects a 2 Ω and an 8 Ω resistor in series to a 10 V battery. (a) Calculate the total resistance of the circuit. (b) Find the current flowing through the circuit. (c) Find the potential difference across the 8 Ω resistor.
Generated by claude-sonnet-4-6 · 2026-06-26 01:07 · grounding rag
Model Answer
(a) Total resistance:
Since the resistors are in series:
$$R_s = R_1 + R_2 = 2\,\Omega + 8\,\Omega = 10\,\Omega$$
(b) Current through the circuit:
By Ohm's law:
$$I = \frac{V}{R_s} = \frac{10\,\text{V}}{10\,\Omega} = 1\,\text{A}$$
(c) Potential difference across the 8 Ω resistor:
Applying Ohm's law to the 8 Ω resistor:
$$V_2 = I \times R_2 = 1\,\text{A} \times 8\,\Omega = 8\,\text{V}$$
Source: Chapter 11, Section 11.6.1 — Resistors in Series
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Explanation
- Examiner focus: Each sub-part carries 1 mark. Show the formula, substitution, and result clearly — don't just write the answer.
- For series circuits: $R_s = R_1 + R_2$ (resistances add up).
- Current is the same throughout a series circuit, so the same $I = 1$ A flows through both resistors.
- For the voltage across one resistor, apply Ohm's law to that resistor alone: $V = IR$.
- Note that $V_1 + V_2 = 2\,\text{V} + 8\,\text{V} = 10\,\text{V}$ = battery voltage ✓ (good self-check).