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A-Level Physics – Electricity

Current, resistance, potential dividers, EMF and internal resistance, capacitors.

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12 questions  ·  40 marks  ·  answers included  ·  A-Level Physics (Edexcel / AQA / OCR)
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This worksheet covers:

  • This worksheet covers current
  • Resistance
  • Potential dividers
  • EMF and internal resistance
  • Capacitors

A calculator is required throughout.

Questions

Section A – Current and Resistance

Skills
1

Define electric current in terms of charge flow, giving the appropriate equation and units.

Show answer
Current is the rate of flow of charge: $I=\dfrac{\Delta Q}{\Delta t}$, where $I$ is in amperes (A), $Q$ is charge in coulombs (C) and $t$ is time in seconds (s).
[2]
2

A charge of $24\,\text{C}$ flows past a point in a circuit in $6\,\text{s}$. Calculate the current.

Show answer
$I=\dfrac{Q}{t}=\dfrac{24}{6}=4\,\text{A}$.
[2]
3

Two resistors of $6\,\Omega$ and $12\,\Omega$ are connected:(a) in series(b) in parallel.Calculate the combined resistance in each case.

Show answer
(a) $R=6+12=18\,\Omega$.(b) $\dfrac{1}{R}=\dfrac{1}{6}+\dfrac{1}{12}=\dfrac{1}{4}$, so $R=4\,\Omega$.
[4]
4

A resistor of resistance $15\,\Omega$ has a potential difference of $6\,\text{V}$ across it. Calculate the current through it and the power dissipated.

Show answer
$I=\dfrac{V}{R}=\dfrac{6}{15}=0.4\,\text{A}$. $P=VI=6\times0.4=2.4\,\text{W}$.
[3]

Section B – EMF, Internal Resistance and Potential Dividers

Exam-style
5

A battery of EMF $9.0\,\text{V}$ and internal resistance $0.5\,\Omega$ is connected to an external resistor of $4.0\,\Omega$.(a) Calculate the current in the circuit.(b) Calculate the terminal potential difference across the battery.

Show answer
(a) $I=\dfrac{\varepsilon}{R+r}=\dfrac{9.0}{4.0+0.5}=2.0\,\text{A}$.(b) Terminal p.d. $=\varepsilon-Ir=9.0-(2.0\times0.5)=8.0\,\text{V}$ (equivalently $V=IR=2.0\times4.0=8.0\,\text{V}$).
[5]
6

Explain why the terminal potential difference of a battery decreases as the current drawn from it increases.

Show answer
Since $V=\varepsilon-Ir$, some of the EMF is used to drive current through the battery’s own internal resistance (the “lost volts”, $Ir$); as $I$ increases, $Ir$ increases, so the terminal p.d. available to the external circuit decreases.
[2]
7

A potential divider consists of two resistors, $2.0\,\text{k}\Omega$ and $8.0\,\text{k}\Omega$, connected in series across a $12\,\text{V}$ supply. Calculate the potential difference across the $8.0\,\text{k}\Omega$ resistor.

Show answer
$V_{\text{out}}=V_{\text{in}}\times\dfrac{R_2}{R_1+R_2}=12\times\dfrac{8.0}{2.0+8.0}=9.6\,\text{V}$.
[3]
8

A light-dependent resistor (LDR) is used with a fixed resistor of $5.0\,\text{k}\Omega$ in a potential divider across a $6.0\,\text{V}$ supply, with the output voltage taken across the fixed resistor. In darkness the LDR resistance is $200\,\text{k}\Omega$; in bright light it is $1.0\,\text{k}\Omega$. Calculate the output voltage in:(a) darkness(b) bright light.

Show answer
(a) $V_{\text{out}}=6.0\times\dfrac{5.0}{5.0+200}=0.146\,\text{V}$ (3 s.f.).(b) $V_{\text{out}}=6.0\times\dfrac{5.0}{5.0+1.0}=5.0\,\text{V}$.
[4]

Section C – Capacitors and Extended Problems

9

A capacitor of capacitance $470\,\mu\text{F}$ is charged to a potential difference of $12\,\text{V}$.(a) Calculate the charge stored.(b) Calculate the energy stored in the capacitor.

Show answer
(a) $Q=CV=470\times10^{-6}\times12=5.64\times10^{-3}\,\text{C}$.(b) $E=\frac{1}{2}CV^2=\frac{1}{2}\times470\times10^{-6}\times12^2=3.38\times10^{-2}\,\text{J}$ (3 s.f.).
[4]
10

A capacitor of capacitance $220\,\mu\text{F}$ discharges through a resistor of $100\,\text{k}\Omega$.(a) Calculate the time constant of the circuit.(b) State what is meant by the “time constant” of a capacitor discharge circuit.

Show answer
(a) $\tau=RC=100\,000\times220\times10^{-6}=22\,\text{s}$.(b) The time constant is the time taken for the charge (or p.d., or current) on a discharging capacitor to fall to $1/e$ (about 37%) of its initial value.
[4]
11

Two resistors, $3.0\,\Omega$ and $6.0\,\Omega$, are connected in parallel, and this combination is connected in series with a $2.0\,\Omega$ resistor and a battery of EMF $12\,\text{V}$ with negligible internal resistance. Calculate the current supplied by the battery.

Show answer
Parallel combination: $\dfrac{1}{R_p}=\dfrac{1}{3.0}+\dfrac{1}{6.0}=\dfrac{1}{2.0}$, so $R_p=2.0\,\Omega$. Total resistance $=2.0+2.0=4.0\,\Omega$. $I=\dfrac{\varepsilon}{R}=\dfrac{12}{4.0}=3.0\,\text{A}$.
[4]
12

A student sets up a circuit to determine the internal resistance of a cell by measuring terminal p.d. for several values of current. Explain how a graph of terminal p.d. (y-axis) against current (x-axis) can be used to find both the EMF and the internal resistance of the cell.

Show answer
Since $V=\varepsilon-Ir$, a graph of $V$ against $I$ is a straight line with $y$-intercept equal to the EMF $\varepsilon$ (the p.d. when $I=0$) and gradient equal to $-r$, so the magnitude of the gradient gives the internal resistance $r$.
[3]

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