Capacitor networks — worked revision

Q40 · Q41 · Q42 · Q43
Q · 40

Charge stored in C

Question

In the circuit shown in figure \(C = 6\,\mu F\). The charge stored in the capacitor of capacity \(C\) is
  • a) zero
  • b) 90 µC
  • c) 40 µC
  • d) 60 µC

The Circuit

C 10 V 2C

Both ends are earthed (0 V), so \(C\) and \(2C\) are simply in series across the 10 V source.

Method & Steps

  1. Series: \(\dfrac{Q}{C}+\dfrac{Q}{2C}=10\)
  2. \(Q\left(\dfrac16+\dfrac1{12}\right)=10\)
  3. \(Q\cdot\dfrac14 = 10\)
  4. \(Q = 40\,\mu C\)

Easy Trick

Series \(C\) and \(2C\) give \(C_{eq}=\tfrac{2C}{3}=4\,\mu F\); \(Q=C_{eq}V=4\times10=40\,\mu C\), the same on both. The “zero” option traps you into thinking the two grounds short it out.

c
\(Q_C = 40\ \mu C\)
Q · 41

Charge stored in the 6 µF

Question

In the circuit shown in figure. Charge stored in 6 µF capacitor will be
  • a) 18 µC
  • b) 54 µC
  • c) 36 µC
  • d) 72 µC

The Circuit

6 µF 4 µF 9 V 12 V

The 9 V source pins the top-right node to 9 V; the 12 V source pins the left rail (= 6 µF’s far plate) to 12 V. The 6 µF therefore only feels the difference.

Method & Steps

  1. 9 V battery → top node at 9 V
  2. 12 V battery → left plate of 6 µF at 12 V
  3. \(V_{6\mu F}=12-9 = 3\,V\)
  4. \(Q = 6\times3 = 18\,\mu C\)

Easy Trick

Two ideal sources clamp the two plates of the 6 µF at fixed potentials (12 V and 9 V). The capacitor sees the gap between them, \(12-9 = 3\,V\) → \(Q = 18\,\mu C\). The other options assume it sees a full battery voltage.

a
\(Q_{6\,\mu F} = 18\ \mu C\)
Q · 42

Potential at B

Question

In the given circuit, if point \(C\) is connected to the earth and a potential of \(+2000\ V\) is given to the point \(A\), the potential at \(B\) is
  • a) 1500 V
  • b) 1000 V
  • c) 500 V
  • d) 400 V

The Circuit

5 µF 10 µF 10 µF 10 µF A B C

A at 2000 V, C at 0 V. From B to C: two 10 µF in series (= 5 µF) in parallel with a single 10 µF, giving 15 µF — and that sits in series with the 5 µF from A to B.

Method & Steps

  1. Top B→C: \(10\) series \(10 = 5\,\mu F\)
  2. \(\parallel\) bottom 10 µF → \(15\,\mu F\)
  3. \(C_{eq}=\dfrac{5\times15}{5+15}=3.75\,\mu F\)
  4. \(Q=3.75\times2000=7500\,\mu C\)
  5. \(V_{AB}=\dfrac{7500}{5}=1500\,V\Rightarrow V_B = 2000-1500 = 500\,V\)

Easy Trick

Series split \(5\,\mu F : 15\,\mu F\) → voltage split \(3:1\). Of 2000 V the 5 µF (A–B) takes \(\tfrac34 = 1500\,V\), so B is \(2000-1500 = 500\,V\) above earth.

c
\(V_B = 500\ V\)
Q · 43

Resultant between A and B

Question

In the given network capacitance \(C_1 = 10\,\mu F\), \(C_2 = 5\,\mu F\) and \(C_3 = 4\,\mu F\). What is the resultant capacitance between \(A\) and \(B\)?
  • a) 2.2 µF
  • b) 3.2 µF
  • c) 1.2 µF
  • d) 4.7 µF

The Circuit

C₁ C₂ C₃ A B

\(C_1\) and \(C_2\) share the top rail and the middle rail → they are in parallel. That pair is then in series with \(C_3\) down to \(B\).

Method & Steps

  1. \(C_1\parallel C_2 = 10+5 = 15\,\mu F\)
  2. Series with \(C_3\): \(\dfrac{15\times4}{15+4}\)
  3. \(=\dfrac{60}{19}\)
  4. \(\approx 3.2\,\mu F\)

Easy Trick

Add the parallel pair first (15 µF), then series with the smaller 4 µF. A series result is always below the smaller capacitor, so it must be under 4 µF → 3.2 µF fits; 4.7 µF is impossible.

b
\(C_{AB} \approx 3.2\ \mu F\)