Science (086) — AI-generated practice question

Derivation:

In a series combination, the same current $I$ flows through each resistor. The potential differences across R₁, R₂, R₃ are V₁, V₂, V₃ respectively.

The total potential difference:
$$V = V_1 + V_2 + V_3 \tag{1}$$

Applying Ohm's law to each resistor:
$$V_1 = IR_1, \quad V_2 = IR_2, \quad V_3 = IR_3$$

If $R_s$ is the equivalent resistance, then $V = IR_s$. Substituting in (1):
$$IR_s = IR_1 + IR_2 + IR_3$$
$$\boxed{R_s = R_1 + R_2 + R_3}$$

Why $R_s$ is always greater than any individual resistance:

Since R₁, R₂, and R₃ are all positive values, their sum $R_s = R_1 + R_2 + R_3$ is necessarily greater than each individual resistance. For example, $R_s > R_1$ because $R_2 + R_3 > 0$.

Source: Chapter 11, Section 11.6.1 – Resistors in Series

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• The two key facts for series are: same current through all resistors, and voltages add up. State both clearly before deriving.
• The derivation follows: apply Ohm's law to the whole circuit ($V = IR_s$) and to each resistor, then substitute into $V = V_1 + V_2 + V_3$.
• For the explanation part (worth ~1 mark), simply state that all resistances are positive, so adding them always gives a sum larger than any single term. One line is enough.