Electric Circuits and Resistance: Question 3

Worked solution

Part (a): Showing the combined resistance

For resistors in parallel, the reciprocal of the combined resistance is the sum of the reciprocals of each resistor:

$\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2}$

Substitute $R_1 = 6\,\Omega$ and $R_2 = 3\,\Omega$:

$\frac{1}{R_{\text{total}}} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$

Taking the reciprocal of both sides:

$R_{\text{total}} = \boxed{2.0\,\Omega}$

as required.

Part (b): Total current from the battery

The p.d. across the combined resistance is the full e.m.f. of the battery, since it has negligible internal resistance. Use $I = V/R$ with the combined resistance from part (a):

$I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{4.0\text{ V}}{2.0\,\Omega}$

$I_{\text{total}} = \boxed{2.0\text{ A}}$

Part (c): Current through $R_2$

In a parallel circuit, the p.d. across each branch is the same as the p.d. across the combination, here, $4.0\text{ V}$. Apply $I = V/R$ to the branch containing $R_2$:

$I_2 = \frac{V}{R_2} = \frac{4.0\text{ V}}{3\,\Omega} = \boxed{1.33\text{ A}}$

(For comparison, the current through $R_1$ is $I_1 = 4.0 \div 6 = 0.67\text{ A}$, and $I_1 + I_2 = 0.67 + 1.33 = 2.0\text{ A}$, matching the total current from part (b), as it must.)

The current through $R_2$ is larger than the current through $R_1$ because the p.d. across both branches is the same, but $R_2$ has a smaller resistance than $R_1$. Since $I = V/R$, a smaller resistance with the same p.d. produces a larger current.

Final answers

• (a) $R_{\text{total}} = \dfrac{1}{\frac{1}{6}+\frac{1}{3}} = \boxed{2.0\,\Omega}$ (shown)
• (b) Total current $=$ $2.0\text{ A}$
• (c) Current through $R_2$ $=$ $1.33\text{ A}$, larger than through $R_1$ because the same p.d. acts across a smaller resistance