Capacitor networks — worked revision

Q36 · Q37 · Q38 · Q39
Q · 36

Charge on the 5 µF

Question

A circuit is shown in the given figure. Find out the charge on the condenser having capacity 5 µF.
  • a) 4.5 µC
  • b) 9 µC
  • c) 7 µC
  • d) 30 µC

The Circuit

3 µF 2 µF 5 µF 4 µF 6 V A B

The battery fixes \(V_{AB}=6\,V\) across every branch. Only the top branch (3 µF in series with \(2\!\parallel\!5\)) decides the 5 µF charge; the 4 µF just sits at 6 V and is irrelevant here.

Method & Steps

  1. \(2\parallel5 = 7\,\mu F\)
  2. Top branch: 3 µF series 7 µF, across 6 V
  3. \(V_{7} = \dfrac{3}{3+7}\times6 = 1.8\,V\)
  4. \(Q_{5} = 5\times1.8 = 9\,\mu C\)

Easy Trick

Series 3 µF & 7 µF split 6 V inversely: \(7\) gets \(\tfrac{3}{10}\times6 = 1.8\,V\). Both 2 µF and 5 µF share that 1.8 V; the 5 µF holds \(5\times1.8 = 9\,\mu C\). The 4 µF is a distractor.

b
\(Q_{5\,\mu F} = 9\ \mu C\)
Q · 37

PD across capacitor B

Question

A potential of \(V = 3000\ V\) is applied to a combination of four initially uncharged capacitors as shown in the figure. Capacitors \(A, B, C\) and \(D\) have capacitances \(C_A = 6.0\,\mu F\), \(C_B = 5.2\,\mu F\), \(C_C = 1.5\,\mu F\) and \(C_D = 3.8\,\mu F\), respectively. If the battery is disconnected, then potential difference across capacitor \(B\) is (approximately)
  • a) 3000 V
  • b) zero
  • c) 530 V
  • d) 350 V

The Circuit

A B D C V

\(A\), the parallel pair \(B\!\parallel\!D\), and \(C\) sit in series across 3000 V. The series charge is common; \(B\) shares the voltage of its parallel block.

Method & Steps

  1. \(B\parallel D = 5.2+3.8 = 9.0\,\mu F\)
  2. \(\dfrac1{C_{eq}}=\dfrac16+\dfrac19+\dfrac1{1.5}\approx0.944\)
  3. \(C_{eq}\approx1.06\,\mu F\)
  4. \(Q=C_{eq}V\approx3176\,\mu C\)
  5. \(V_B=\dfrac{Q}{9}\approx 350\,V\)

Easy Trick

B and D are in parallel, so \(V_B = V_D = \) the block voltage. Get the series charge once, divide by the block’s 9 µF → ~350 V. (Disconnecting the battery doesn’t change anything — the charges are already set.)

d
\(V_B \approx 350\ V\)
Q · 38

Voltage across RS

Question

Four capacitors are arranged as shown. All are initially uncharged. A 30 V battery is placed across terminal \(PQ\) to charge the capacitors and is then removed. The voltage across the terminals \(RS\) is then (in volt)
  • a) 10
  • b) 20
  • c) 30
  • d) 40

The Circuit

1 µF 3 µF 3 µF 3 µF P Q R S

The 1 µF sits straight across \(PQ\). The three 3 µF form a chain \(P\!\to\!R\!\to\!S\!\to\!Q\) — three equal capacitors in series across the same 30 V.

Method & Steps

  1. \(V_{PQ}=30\,V\) (battery)
  2. Chain P→R→S→Q: three equal 3 µF in series
  3. Each takes \(\dfrac{30}{3}=10\,V\)
  4. \(V_{RS}=V_R-V_S = 10\,V\)

Easy Trick

Equal capacitors in series split the voltage equally. \(R\) and \(S\) are the middle nodes of the three-step chain, so the slice between them is exactly one-third of 30 V = 10 V. Removing the battery changes nothing (the loop is already balanced at 30 V).

a
\(V_{RS} = 10\ V\)
Q · 39

Find C from the equivalent

Question

If the equivalent capacitance between points \(P\) and \(Q\) of the combination of the capacitors show in figure below is 30 µF, the capacitor \(C\) is
  • a) 60 µF
  • b) 30 µF
  • c) 10 µF
  • d) 5 µF

The Circuit

C 20 µF 20 µF 20 µF P Q

The three 20 µF share both buses → parallel \((=60\,\mu F)\). That block is in series with \(C\), and the whole thing equals 30 µF.

Method & Steps

  1. Three in parallel: \(3\times20 = 60\,\mu F\)
  2. Series with C: \(\dfrac{60\,C}{60+C}=30\)
  3. \(60C = 30(60+C)\Rightarrow 30C = 1800\)
  4. \(C = 60\,\mu F\)

Easy Trick

Two capacitors in series give exactly half when they’re equal. The target 30 µF is half of 60 µF, so \(C\) must equal the 60 µF block → \(C = 60\,\mu F\).

a
\(C = 60\ \mu F\)