Capacitor networks — worked revision

Q28 · Q29
Q · 28

Potential at the mid-point P

Question

In the figure a potential of \(+1200\ V\) is given to point \(A\) and point \(B\) is earthed, what is the potential at the point \(P\)?
  • a) 100 V
  • b) 200 V
  • c) 400 V
  • d) 800 V

The Circuit

3 µF 4 µF 2 µF A +1200 V P B

A is held at 1200 V, B at 0 V (earthed). The 3 µF carries the full chain charge, and the 4 µF & 2 µF share the node between P and B as a parallel pair.

Concept

Collapse \(4\parallel2 = 6\,\mu F\). Then 3 µF and 6 µF are in series across 1200 V. Same chain charge \(Q\); the potential at P is whatever is left after the drop across the 3 µF.

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. Drop on 3 µF: \(\dfrac{2400}{3}=800\,V\)
  5. \(V_P = 1200-800 = 400\,V\)

Easy Trick

Series voltage splits inversely to capacitance. Here \(3\,\mu F : 6\,\mu F\) → voltage split \(2:1\), so of 1200 V the 3 µF takes 800 V and the parallel block takes 400 V. \(V_P\) = drop across the block from earthed B = 400 V.

c
\(V_P = 400\ V\)
Q · 29

Charge on the 4 µF capacitor

Question

The charge on 4 µF capacitor in the given circuit is (in µC)
  • a) 12
  • b) 24
  • c) 36
  • d) 32

The Circuit

4 µF 1 µF 5 µF 3 µF 10 V

The top branch (4 µF in series with 1 µF ∥ 5 µF) and the 3 µF branch both connect the same two nodes that the battery drives, so each branch sees the full 10 V.

Concept

Reduce the top branch alone — the 3 µF branch doesn’t affect it. \(1\parallel5 = 6\,\mu F\); in series with 4 µF gives the branch capacitance. That whole branch sits across 10 V, and a series branch carries one common charge — which is the charge on the 4 µF.

Method & Steps

  1. \(1\parallel5 = 6\,\mu F\)
  2. Top branch: \(\dfrac{4\times6}{4+6}=2.4\,\mu F\)
  3. Across 10 V: \(Q = 2.4\times10\)
  4. \(Q_{4\mu F}= 24\,\mu C\)

Easy Trick

Ignore the parallel 3 µF branch — it shares the same 10 V but carries its own separate charge. For the 4 µF, just reduce its own series branch (4 with \(1\!\parallel\!5=6\)) to 2.4 µF and multiply by 10 V → 24 µC. Series branch ⇒ that’s the charge on every element in it.

b
\(Q_{4\,\mu F} = 24\ \mu C\)