Electric Circuits and Resistance: Question 7

Worked solution

Part (a): Potential difference across $R_1$

Use $V = IR$ with the current and the resistance of $R_1$:

$V_1 = I \times R_1 = 0.20\text{ A} \times 15\,\Omega$

$V_1 = \boxed{3.0\text{ V}}$

Part (b): Potential difference across X

In a series circuit, the potential differences across the individual components add up to the e.m.f. of the source (since the battery has negligible internal resistance, none of the e.m.f. is “lost” elsewhere):

$\varepsilon = V_1 + V_X$

Rearrange and substitute $\varepsilon = 10\text{ V}$ and $V_1 = 3.0\text{ V}$ from part (a):

$V_X = \varepsilon - V_1 = 10\text{ V} - 3.0\text{ V}$

$V_X = \boxed{7.0\text{ V}}$

Part (c): Resistance of X

Use $R = V/I$ with the p.d. across X found in part (b) and the current, which is the same everywhere in a series circuit:

$R_X = \frac{V_X}{I} = \frac{7.0\text{ V}}{0.20\text{ A}}$

$R_X = \boxed{35\,\Omega}$

(As a check: the combined resistance of the circuit is $R_1 + R_X = 15 + 35 = 50\,\Omega$, and $\varepsilon \div R_{\text{total}} = 10 \div 50 = 0.20\text{ A}$, matching the current given in the question.)

Final answers

• (a) P.d. across $R_1$ $=$ $3.0\text{ V}$
• (b) P.d. across X $=$ $7.0\text{ V}$ (since the p.d.s across components in series add up to the e.m.f.)
• (c) Resistance of X $=$ $35\,\Omega$