Q · 25

Four-capacitor network — charge & PD across the 4 µF

Capacitor networks

Question

In the network shown, four capacitors are connected across a 300 V supply. Find the charge and the potential difference across each 4 µF capacitor.

The Circuit

300 V 20 µF 12 µF 4 µF 4 µF

The two 4 µF branches share the same two nodes ⇒ they are in parallel (\(=8\,\mu F\)). That pair sits in series with the 20 µF and 12 µF, and the 300 V is across the whole chain.

Given

Supply\(V = 300\) V
Top branch\(20\ \mu F\)
Bottom branch\(12\ \mu F\)
Middle pair\(4\ \mu F \parallel 4\ \mu F\)

Asked

Charge on one 4 µF\(q_4 = ?\)
PD across one 4 µF\(V_4 = ?\)

Concept

Identify parallel groups first, then collapse the series chain. In a series chain the same charge \(Q\) flows through every element, so the parallel block carries the full \(Q\); the two equal 4 µF capacitors then split that charge but share the same voltage.

$$ \text{Parallel: } C_p = 4+4 = 8\ \mu F \qquad \text{Series: } \frac1{C_{eq}}=\frac1{20}+\frac1{8}+\frac1{12} $$

Method & Steps

  1. Combine the two 4 µF in parallel: \(C_p = 8\ \mu F.\)
  2. Series of 20, 8, 12 µF: \(\dfrac1{C_{eq}}=\dfrac{6+15+10}{120}=\dfrac{31}{120}\Rightarrow C_{eq}=\dfrac{120}{31}\approx 3.87\ \mu F.\)
  3. Series charge (same everywhere): \(Q = C_{eq}V = \dfrac{120}{31}\times 300 = \dfrac{36000}{31}\approx 1161\ \mu C.\)
  4. PD across the 8 µF block: \(V_p = \dfrac{Q}{C_p}=\dfrac{1161}{8}\approx 145\ V.\)
  5. Each 4 µF has the same \(V_4 = 145\ V\), so \(q_4 = C\,V_4 = 4\times 145 \approx 580\ \mu C.\)

Easy Trick

Equal parallel capacitors split charge equally but keep one shared voltage. Find \(Q\) through the chain, divide by the parallel 8 µF to get \(V_4\approx145\) V, then each 4 µF carries \(\tfrac{Q}{2}\approx 580\) µC. The total \(Q\approx1161\) µC = \(2\times 580\) — a quick self-check.

d
Across each 4 µF: charge \(\approx 580\ \mu C\), PD \(\approx 145\ V\).
Both 4 µF share 145 V (parallel); together they hold the full series charge of ~1161 µC.