Electric Circuits and Resistance: Question 9

Worked solution

Part (a): Combined resistance of $R_2$ and $R_3$

$R_2$ and $R_3$ are in parallel, so their combined resistance $R_{23}$ satisfies:

$\frac{1}{R_{23}} = \frac{1}{R_2} + \frac{1}{R_3}$

Substitute $R_2 = 6.0\,\Omega$ and $R_3 = 12\,\Omega$:

$\frac{1}{R_{23}} = \frac{1}{6.0} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$

Taking the reciprocal of both sides:

$R_{23} = \boxed{4.0\,\Omega}$

Part (b): Total resistance of the circuit

$R_1$ is in series with the parallel combination, so their resistances simply add:

$R_{\text{total}} = R_1 + R_{23} = 4.0\,\Omega + 4.0\,\Omega$

$R_{\text{total}} = \boxed{8.0\,\Omega}$

Part (c): Total current from the battery

The full e.m.f. of the battery acts across the total resistance of the circuit, since the battery has negligible internal resistance. Use $I = V/R$:

$I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{8.0\,\Omega}$

$I_{\text{total}} = \boxed{1.5\text{ A}}$

Part (d): P.d. across the parallel combination

The current found in part (c) flows through $R_1$ and then through the parallel combination, since $R_1$ and the parallel section are in series with each other and carry the same total current. Use $V = IR$ with the combined resistance from part (a):

$V_{23} = I_{\text{total}} \times R_{23} = 1.5\text{ A} \times 4.0\,\Omega$

$V_{23} = \boxed{6.0\text{ V}}$

(As a check: the p.d. across $R_1$ is $1.5 \times 4.0 = 6.0\text{ V}$, and $6.0 + 6.0 = 12\text{ V}$, matching the e.m.f. of the battery, as it must. Also, the branch currents are $I_2 = 6.0 \div 6.0 = 1.0\text{ A}$ and $I_3 = 6.0 \div 12 = 0.5\text{ A}$, which sum to $1.0 + 0.5 = 1.5\text{ A}$, matching the total current found in part (c).)

Final answers

• (a) $R_{23} = \boxed{4.0\,\Omega}$
• (b) Total resistance $=$ $8.0\,\Omega$
• (c) Total current $=$ $1.5\text{ A}$
• (d) P.d. across the parallel combination $=$ $6.0\text{ V}$