Worked examples · Sheet 3

Capacitors in series and in parallel

Theory used: Capacitors in parallel Capacitors in series Capacitor

Exercise 1

Two capacitors of capacitances $C_{1} = 2.0 μ F$ and $C_{2} = 3.0 μ F$ , initially uncharged, are connected to a $Δ V = 12 V$ source, first in parallel and then, after being discharged, in series. In both cases compute the equivalent capacitance, the charge and the potential difference of each capacitor.

parallel: C_{eq} = C_{1} + C_{2}, Δ V_{1} = Δ V_{2} = Δ V, Q_{i} = C_{i} Δ V

series: C_{eq} = \frac{C _{1} C _{2}}{C _{1} + C _{2}}, Q_{1} = Q_{2} = C_{eq} Δ V, Δ V_{i} = \frac{Q _{1}}{C _{i}}

parallel: C_{eq} series: C_{eq} Δ V_{1} = 5.0 μ F, Q_{1} = 2.0 \cdot 12 μ C = 24 μ C, Q_{2} = 3.0 \cdot 12 μ C = 36 μ C = \frac{2.0 \cdot 3.0}{5.0} μ F = 1.2 μ F, Q_{1} = Q_{2} = 1.2 \cdot 12 μ C \approx 14 μ C = \frac{14.4}{2.0} V = 7.2 V, Δ V_{2} = \frac{14.4}{3.0} V = 4.8 V

In parallel the plates connected to the same wire are at the same potential, therefore both capacitors have the potential difference of the source, and by the theorem on capacitors in parallel the capacitances add. The charge of each is proportional to its capacitance, and the total charge $60 μ C$ equals $C_{eq} Δ V$ .

In series the inner plates form an isolated neutral conductor, hence the two capacitors have the same charge, and by the theorem on capacitors in series the reciprocals of the capacitances add. The equivalent capacitance, $1.2 μ F$ , is smaller than both. The capacitor with the smaller capacitance has the larger voltage, and the two voltages add up to $12 V$ , as required by the sum of the potential differences along the series.

Exercise 2

In the network in the figure $C_{1} = 4.0 μ F$ is connected in series with the parallel combination of $C_{2} = 2.0 μ F$ and $C_{3} = 6.0 μ F$ , and the network is connected to a $Δ V = 24 V$ source. The capacitors are initially uncharged. Compute the equivalent capacitance, the charge and the potential difference of each capacitor.

Capacitor network:

C_{1}

is in series with the parallel combination of

C_{2}

and

C_{3}

C_{23} = C_{2} + C_{3}, C_{eq} = \frac{C _{1} C _{23}}{C _{1} + C _{23}}, Q_{1} = C_{eq} Δ V

Δ V_{1} = \frac{Q _{1}}{C _{1}}, Δ V_{23} = \frac{Q _{1}}{C _{23}}, Q_{2} = C_{2} Δ V_{23}, Q_{3} = C_{3} Δ V_{23}

C_{23} C_{eq} Q_{1} Δ V_{1} Q_{2} = (2.0 + 6.0) μ F = 8.0 μ F = \frac{4.0 \cdot 8.0}{12} μ F \approx 2.7 μ F = \frac{8}{3} \cdot 24 μ C = 64 μ C = \frac{64}{4.0} V = 16 V, Δ V_{23} = \frac{64}{8.0} V = 8.0 V = 2.0 \cdot 8.0 μ C = 16 μ C, Q_{3} = 6.0 \cdot 8.0 μ C = 48 μ C

$C_{2}$ and $C_{3}$ have their plates connected to the same two nodes, therefore they are in parallel and are replaced by $C_{23} = 8.0 μ F$ . This equivalent capacitor is in series with $C_{1}$ , because between the two there is a single wire, which does not lead to other components; the capacitance of the network is hence $\frac{8}{3} μ F$ .

The network is then traversed backward. The charge delivered by the source is that of the equivalent capacitor, and because of the series connection it is also the charge of $C_{1}$ and of the parallel combination. From the charge the voltages are obtained, which add up to $24 V$ . Finally $C_{2}$ and $C_{3}$ have the same voltage $8.0 V$ , and their charges, $16 μ C$ and $48 μ C$ , add up again to $64 μ C$ , as a check.

Exercise 3

Two capacitors of capacitances $C_{1} = 1.0 μ F$ and $C_{2} = 2.0 μ F$ can each withstand a maximum potential difference of $100 V$ . Compute the maximum potential difference that can be applied to the two capacitors connected in series, and the one that can be applied to their parallel connection.

Q_{1} = Q_{2} = Q, Δ V_{1} = \frac{Q}{C _{1}}, Δ V_{2} = \frac{Q}{C _{2}} = \frac{C _{1}}{C _{2}} Δ V_{1}

Δ V_{1} Δ V_{2} Δ V_{m a x}^{series} = 100 V ⟹ Q = C_{1} \cdot 100 V = 100 μ C = \frac{100}{2.0} V = 50 V = Δ V_{1} + Δ V_{2} = 150 V, Δ V_{m a x}^{parallel} = 100 V

In series the two capacitors have the same charge, therefore the larger voltage belongs to the capacitor with the smaller capacitance, $C_{1}$ , and at every instant $Δ V_{1} = 2 Δ V_{2}$ . As the applied voltage increases, the first capacitor to reach the limit is hence $C_{1}$ , and this happens with the charge $Q = 100 μ C$ , when $C_{2}$ is at $50 V$ . The maximum voltage that can be applied to the series is the sum, $150 V$ ; as a check, $C_{eq} = \frac{2}{3} μ F$ and $C_{eq} \cdot 150 V = 100 μ C$ .

The result is smaller than $200 V$ , the sum of the maximum voltages, because the voltages are shared unequally. In parallel both capacitors have the applied voltage, therefore the limit is $100 V$ .