B.5 Current and circuits

Practice Worksheet — name: ______________________ date: ____________

Formulas for this topic

Electric current

I = \frac{\Delta q}{\Delta t}

Potential difference

V = \frac{W}{q}

Resistance

R = \frac{V}{I}

Resistivity

\rho = \frac{RA}{L}

Electrical power

P = VI = I^{2}R = \frac{V^{2}}{R}

Resistors in series

R_{s} = R_{1} + R_{2} + \ldots

Resistors in parallel

\frac{1}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \ldots

EMF and internal resistance

\varepsilon = I(R + r)

Section A — Multiple choice

A1. Two identical resistors connected in parallel have combined resistance $R_p$ . Connected in series, their combined resistance is:

A $R_p$
B $2R_p$
C $4R_p$
D $R_p/4$

A2. A battery of emf 12 V and internal resistance $0.50\ \Omega$ delivers a current of 4.0 A. Its terminal potential difference is:

A 12 V
B 10 V
C 14 V
D 8.0 V

A3. The current through a filament lamp doubles when the potential difference across it is tripled. The lamp is:

A Ohmic, because current increases with voltage
B Non-ohmic, because resistance has increased with temperature
C Non-ohmic, because resistance has decreased with temperature
D Ohmic, because the filament obeys $P = VI$

Section B — Short answer

B1. A copper wire of length 2.0 m and cross-sectional area $1.0\times10^{-6}\ \text{m}^2$ has resistivity $1.7\times10^{-8}\ \Omega\text{m}$ . Calculate its resistance, and the new resistance if the wire is stretched (at constant volume) to twice its length. [4 marks]

B2. Explain why the emf of a cell exceeds its terminal potential difference when a current flows, and state when they are equal. [3 marks]

B3. Three resistors — $6.0\ \Omega$ and $3.0\ \Omega$ in parallel, in series with $4.0\ \Omega$ — are connected to a 12 V supply of negligible internal resistance. Find the current drawn from the supply. [3 marks]

Answer key

For worked explanations, interactive practice and more free resources, visit www.newtonine.com

Section A

A1: $4R_p$ — For two equal resistors $R$ : parallel gives $R/2$ , series gives $2R$ . The ratio series/parallel is $2R \div R/2 = 4$ .

A2: 10 V — The internal resistance drops $Ir = 4.0 \times 0.50 = 2.0\ \text{V}$ inside the battery, leaving $V = \varepsilon - Ir = 10\ \text{V}$ across the terminals. Terminal voltage always sags under load.

A3: Non-ohmic, because resistance has increased with temperature — Ohmic behaviour means $I \propto V$ . Here tripling $V$ only doubles $I$ , so $R = V/I$ has grown — the filament heats up and its lattice ions vibrate more, scattering electrons more effectively.

Section B

B1: $R = \rho L/A = (1.7\times10^{-8})(2.0)/(1.0\times10^{-6}) = 0.034\ \Omega$ . Stretching to $2L$ at constant volume halves $A$ , so $R' = \rho(2L)/(A/2) = 4R = 0.136\ \Omega$ . Resistance scales with the square of the stretch factor.

B2: The emf is the total energy per unit charge supplied by the cell; some of it is dissipated crossing the cell's own internal resistance ( $Ir$ ), so the terminals only receive $V = \varepsilon - Ir$ . They are equal when no current flows (open circuit) or for an ideal cell with $r = 0$ .

B3: Parallel pair: $(1/6 + 1/3)^{-1} = 2.0\ \Omega$ . Total: $2.0 + 4.0 = 6.0\ \Omega$ . Current: $I = 12/6.0 = 2.0\ \text{A}$ .