Electric Circuits and Resistance: Question 8

Worked solution

Part (a): Combined e.m.f. of the two cells

When cells are connected in series, their e.m.f.s add together:

$\varepsilon_{\text{total}} = \varepsilon_1 + \varepsilon_2 = 1.5\text{ V} + 1.5\text{ V}$

$\varepsilon_{\text{total}} = \boxed{3.0\text{ V}}$

Part (b): Current in the resistor

Rearrange $R = V/I$ to make current the subject:

$I = \frac{V}{R}$

The combined e.m.f. drives the current through the single resistor, so use $V = 3.0\text{ V}$ (the combined e.m.f. from part (a)) and $R = 5.0\,\Omega$:

$I = \frac{3.0\text{ V}}{5.0\,\Omega}$

$I = \boxed{0.60\text{ A}}$

Part (c): Potential difference across the resistor

Because the cells and connecting wires have negligible resistance, the resistor is the only component in the circuit where electrical energy is transferred to something other than the charge carriers themselves. All of the energy supplied by the cells per unit charge is therefore transferred to the resistor, so:

$V_{\text{resistor}} = \boxed{3.0\text{ V}}$

This is equal to the combined e.m.f. found in part (a). With no internal resistance and only one resistor in the circuit, the p.d. across that resistor must equal the total e.m.f. of the battery.

Final answers

• (a) Combined e.m.f. $=$ $3.0\text{ V}$
• (b) Current $=$ $0.60\text{ A}$
• (c) P.d. across the resistor $=$ $3.0\text{ V}$, equal to the combined e.m.f.