AS Level Physics 9702: Potential Difference, Charge and Power in a Resistor

A battery maintains a constant potential difference of $9.00\text{ V}$ across a resistor of resistance $15.0\,\Omega$ . The circuit is switched on for $4.00$ minutes.

(a) State what is meant by potential difference. [1]

(b) Calculate the current $I$ in the resistor. [2]

(d) The charge on a single electron is $e = 1.60\times10^{-19}\text{ C}$ . Use $Q = ne$ to calculate the number $n$ of electrons that flow through the resistor in this time. [2]

(e) Calculate the power $P$ dissipated in the resistor using $P = VI$ , and show that this agrees with the value obtained using $P = I^2R$ . [2]

Part (a): Meaning of potential difference

The potential difference across a component is the (electrical) energy transferred, or work done, per unit charge as charge passes through it: $V = \frac{W}{Q}$

Part (b): Current in the resistor

The battery maintains a constant potential difference $V = 9.00\text{ V}$ across the resistor of resistance $R = 15.0\,\Omega$ . Using Ohm’s law: $I = \frac{V}{R} = \frac{9.00}{15.0} = 0.600\text{ A}$

Check: $R \times I = 15.0 \times 0.600 = 9.00\text{ V}$ , which matches the given potential difference. $\checkmark$

Part (c): Charge that flows

First convert the time to seconds: $t = 4.00\text{ min} = 4.00 \times 60 = 240\text{ s}$

Using $Q = It$ : $Q = 0.600 \times 240 = 144\text{ C}$

Check: $Q / t = 144 / 240 = 0.600\text{ A}$ , which matches the current found in (b). $\checkmark$

Part (d): Number of electrons, using Q = ne

Charge is carried by discrete electrons, so the total charge is related to the number of electrons $n$ and the charge on one electron $e$ by: $Q = ne \implies n = \frac{Q}{e}$

Substituting $Q = 144\text{ C}$ and $e = 1.60\times10^{-19}\text{ C}$ : $n = \frac{144}{1.60\times10^{-19}} = 9.00\times10^{20}$

Check: $n \times e = 9.00\times10^{20} \times 1.60\times10^{-19} = 14.4\times10^{1} = 144\text{ C}$ , which matches $Q$ from (c). $\checkmark$

Part (e): Power dissipated

Using $P = VI$ with $V = 9.00\text{ V}$ and $I = 0.600\text{ A}$ : $P = VI = 9.00 \times 0.600 = 5.40\text{ W}$

Now checking with $P = I^2R$ , using $I = 0.600\text{ A}$ and $R = 15.0\,\Omega$ : $P = I^2R = (0.600)^2 \times 15.0 = 0.360 \times 15.0 = 5.40\text{ W}$

Both routes give $5.40\text{ W}$ , so the two expressions for power agree. (As a further check, $P = V^2/R = 9.00^2/15.0 = 81.0/15.0 = 5.40\text{ W}$ too, confirming that $P = VI = I^2R = V^2/R$ all give the same value here.)

Final answers

(a) Potential difference $=$ energy transferred (work done) per unit charge, $V = W/Q$ .
(b) Current $=$ $0.600\text{ A}$
(c) Charge $=$ $144\text{ C}$
(d) Number of electrons $=$ $9.00\times10^{20}$
(e) Power $=$ $5.40\text{ W}$ , consistent whether found from $P = VI$ or $P = I^2R$ (and also $P = V^2/R$ )