Series capacitors — worked revision

Q26 · Q27
Q · 26

Potential of the junction point

Question

Two condensers \(C_1\) and \(C_2\) in a circuit are joined as shown in figure. The potential of point \(A\) is \(V_1\) and that of \(B\) is \(V_2\). The potential of point \(D\) will be
  • a) \(\dfrac{1}{2}\,(V_1+V_2)\)
  • b) \(\dfrac{C_2V_1+C_1V_2}{C_1+C_2}\)
  • c) \(\dfrac{C_1V_1+C_2V_2}{C_1+C_2}\)
  • d) \(\dfrac{C_2V_1-C_1V_2}{C_1+C_2}\)

The Circuit

C₁ C₂ A V₁ D B V₂

\(C_1\) sits between \(A\) and \(D\); \(C_2\) between \(D\) and \(B\) — a series pair. Node \(D\) is isolated, so the charges meeting there must cancel.

Concept

Charge magnitude is equal across series capacitors. Equating the charge fed in through \(C_1\) to that drawn out through \(C_2\) fixes \(V_D\) — a capacitance-weighted average of \(V_1\) and \(V_2\).

Method & Steps

  1. \(C_1(V_1-V_D)=C_2(V_D-V_2)\)
  2. \(C_1V_1+C_2V_2=(C_1+C_2)V_D\)
  3. \(V_D=\dfrac{C_1V_1+C_2V_2}{C_1+C_2}\)

Easy Trick

A junction between two capacitors sits at the capacitance-weighted average of the two end potentials: each side pulls \(V_D\) toward itself in proportion to its own \(C\). Equal \(C\) → simple midpoint \(\tfrac12(V_1+V_2)\); the bigger capacitor wins more.

c
\(V_D=\dfrac{C_1V_1+C_2V_2}{C_1+C_2}\)
Q · 27

PD across the 6 µF in a series chain

Question

Three capacitors of 2 µF, 3 µF and 6 µF are joined in series and the combination is charged by means of a 24 V battery. The potential difference between the plates of the 6 µF capacitor is
  • a) 4 V
  • b) 6 V
  • c) 8 V
  • d) 10 V

The Circuit

24 V 2 µF 3 µF 6 µF

In series every capacitor carries the same charge \(Q\). Find \(Q\) once from the equivalent capacitance, then read off each voltage as \(Q/C\).

Concept

Series: \(\dfrac1{C_{eq}}=\sum\dfrac1{C_i}\), and the chain charge \(Q=C_{eq}V\). The smallest capacitor takes the largest share of voltage; the largest (6 µF) takes the least.

Method & Steps

  1. \(\dfrac1{C_{eq}}=\dfrac12+\dfrac13+\dfrac16=1\Rightarrow C_{eq}=1\,\mu F\)
  2. \(Q=C_{eq}V=1\times24=24\,\mu C\)
  3. \(V_{6}=\dfrac{Q}{6}=\dfrac{24}{6}=4\,V\)

Easy Trick

Series voltages split inversely to capacitance. Ratio \(2:3:6\) → voltage ratio \(\tfrac12:\tfrac13:\tfrac16 = 3:2:1\), summing to 6 parts of 24 V = 4 V each part. So \(V_6 = 1\times4 = 4\,V\) (and check: \(12+8+4=24\,V\)).

a
\(V_{6\,\mu F} = 4\ V\)