Capacitor networks — worked revision

Q30 · Q35
Q · 30

Bridge of four identical capacitors

Question

Four identical capacitors are connected as shown in diagram. When a battery of 6 V is connected between \(A\) and \(B\), then the charge stored is found to be 1.5 µC. The value of \(C_1\) is
  • a) 2.5 µF
  • b) 15 µF
  • c) 1.5 µF
  • d) 0.1 µF

The Circuit

6 V C₁ C₁ C₁ C₁ A B

Two opposite arms are bare wires, so \(A\) merges with the left node and \(B\) with the right node. All four capacitors then bridge the same pair \(A\)–\(B\).

Concept

After the merge: the two side capacitors are each \(C_1\) directly across \(A\)–\(B\); the two middle ones are in series \((=C_1/2)\), also across \(A\)–\(B\). All three paths are in parallel.

Method & Steps

  1. Two side caps: \(C_1 + C_1\)
  2. Two middle in series: \(\tfrac{C_1}{2}\)
  3. \(C_{eq}=C_1+C_1+\tfrac{C_1}{2}=2.5\,C_1\)
  4. \(Q=C_{eq}V:\;1.5=2.5\,C_1(6)\)
  5. \(C_1 = \dfrac{1.5}{15}=0.1\,\mu F\)

Easy Trick

Spot the bare-wire arms first — they collapse the diamond so every capacitor lands across \(A\)–\(B\). Then it’s just \(C_1+C_1+\tfrac{C_1}{2}=2.5\,C_1\). With \(C_{eq}=Q/V = 1.5/6 = 0.25\,\mu F\), divide by 2.5 → \(0.1\,\mu F\).

d
\(C_1 = 0.1\ \mu F\)
Q · 35

Charge on the 4 µF

Question

In the given circuit, if point \(b\) is connected to earth and a potential of 1200 V is given to a point \(a\), the charge on 4 µF capacitor is
  • a) 800 µC
  • b) 1600 µC
  • c) 2400 µC
  • d) 3000 µC

The Circuit

3 µF 4 µF 2 µF a 1200 V b

\(a\) at 1200 V, \(b\) at 0 V. The 3 µF carries the whole chain charge; the 4 µF and 2 µF form a parallel pair between the mid-node and earthed \(b\).

Concept

Collapse \(4\parallel2 = 6\,\mu F\); series with 3 µF gives \(C_{eq}\) and the chain charge \(Q\). The block’s voltage is \(Q/6\); the 4 µF (inside the block) then holds \(4\times V_{block}\).

Method & Steps

  1. \(4\parallel2 = 6\,\mu F\)
  2. \(C_{eq}=\dfrac{3\times6}{3+6}=2\,\mu F\)
  3. \(Q=2\times1200=2400\,\mu C\)
  4. \(V_{block}=\dfrac{2400}{6}=400\,V\)
  5. \(Q_{4\mu F}=4\times400=1600\,\mu C\)

Easy Trick

Series split \(3\,\mu F:6\,\mu F\) → voltage split \(2:1\), so the 6 µF block gets \(\tfrac13\times1200 = 400\,V\). In a parallel block, charge follows capacitance, so the 4 µF takes \(\tfrac{4}{6}\) of the \(2400\,\mu C\) total → 1600 µC (and the 2 µF takes 800 µC).

b
\(Q_{4\,\mu F} = 1600\ \mu C\)