D.C. Circuits: Question 2

Worked solution

Part (a): Combined resistance of the parallel pair

For two resistors in parallel, derived from Kirchhoff’s laws: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6.0} + \frac{1}{12.0}$

Writing both fractions over a common denominator of $12$: $\frac{1}{R_p} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$

So $R_p = 4.0\ \Omega$.

Check using decimals instead of fractions: $\frac{1}{6.0}=0.1667$ and $\frac{1}{12.0}=0.0833$; adding gives $0.1667+0.0833=0.250=\frac{1}{4.0}$, so $R_p=4.0\ \Omega$. Both routes agree.

Part (b): Total resistance of the network

The $4.0\ \Omega$ resistor is in series with the parallel combination, so the resistances simply add: $R_{\text{total}} = R_3 + R_p = 4.0 + 4.0 = 8.0\ \Omega$

Part (c): Total current from the battery

Since the internal resistance is negligible, the full e.m.f. acts as the p.d. across the whole network: $I = \frac{V}{R_{\text{total}}} = \frac{12}{8.0} = 1.5\text{ A}$

Check: $1.5\times8.0=12\text{ V}$, which matches the battery’s e.m.f. Confirmed.

Part (d): Currents in the parallel branches

First find the p.d. across the parallel section using the total current and $R_p$ from part (a): $V_p = I \times R_p = 1.5 \times 4.0 = 6.0\text{ V}$

Check using the series resistor instead: the p.d. across the $4.0\ \Omega$ resistor is $I\times R_3=1.5\times4.0=6.0\text{ V}$, and $6.0+6.0=12\text{ V}$ matches the battery e.m.f. Confirmed.

This same p.d. of $6.0\text{ V}$ appears across both parallel branches (that is what “parallel” means), so: $I_1 = \frac{V_p}{R_1} = \frac{6.0}{6.0} = 1.0\text{ A} \qquad I_2 = \frac{V_p}{R_2} = \frac{6.0}{12.0} = 0.5\text{ A}$

Kirchhoff’s first law check: the two branch currents should recombine to give the total current entering the parallel section: $I_1 + I_2 = 1.0 + 0.5 = 1.5\text{ A} = I$ This matches the total current found in part (c), confirming both branch currents are correct.

Final answers

• (a) $R_p = \boxed{4.0}\ \Omega$
• (b) $R_{\text{total}} = \boxed{8.0}\ \Omega$
• (c) $I = \boxed{1.5}\text{ A}$
• (d) Current in $6.0\ \Omega$ resistor $=\boxed{1.0}\text{ A}$; current in $12\ \Omega$ resistor $=\boxed{0.5}\text{ A}$; sum $=1.5\text{ A}=I$, consistent with Kirchhoff’s first law