Charge sharing & mixed network — worked revision

Q33 · Q34
Q · 33

Two conductors share charge

Question

Consider two conductors. One of them has a capacity of 2 units and the capacity of the other is unknown. They are charged until their potentials are 4 and 5 units respectively. The two conductors are now connected by a wire when their common potential is found to be 4.6 units. Then the unknown capacity has the value (in the same units as above)
  • a) 6
  • b) 5
  • c) 4
  • d) 3

The Picture — before & after

C = 2 V = 4 C = ? V = 5 separate join wire 4.6 4.6 common V = 4.6

Joining by a wire forces both to one potential, but the total charge is unchanged. That single conservation law pins down the unknown capacity.

Concept

Charge before = charge after: \(C_1V_1+C_2V_2=(C_1+C_2)V_c.\) The common potential is the charge-weighted average of the two starting potentials.

Method & Steps

  1. \(2(4)+C_2(5)=(2+C_2)(4.6)\)
  2. \(8+5C_2 = 9.2+4.6C_2\)
  3. \(0.4\,C_2 = 1.2\)
  4. \(C_2 = 3\)

Easy Trick

The common potential lands closer to whichever conductor holds more charge. 4.6 is \(0.6\) above 4 and \(0.4\) below 5, a \(3:2\) split — so the unknown carries \(3/2\) the charge-influence of the 2-unit one ⇒ \(C_2 = 3\). (Lever rule: \(C_1(V_c-V_1)=C_2(V_2-V_c)\).)

d
Unknown capacity \(= 3\) units
Q · 34

Voltage across the 2 µF

Question

Two capacitors 2 µF and 4 µF are connected in parallel. A third capacitor of 6 µF capacity is connected in series. The combination is connected across a 12 V battery. The voltage across a 2 µF capacitor is
  • a) 2 V
  • b) 6 V
  • c) 8 V
  • d) 1 V

The Circuit

2 µF 4 µF 6 µF 12 V

The 2 µF and 4 µF share the same two nodes → parallel (= 6 µF). That block is in series with the single 6 µF, all driven by 12 V.

Concept

Series chain carries one common charge \(Q\). Find \(Q\) from \(C_{eq}\), then the parallel block’s voltage is \(Q/C_{block}\) — and the 2 µF, being inside that block, shares exactly that voltage.

Method & Steps

  1. \(2\parallel4 = 6\,\mu F\)
  2. \(C_{eq}=\dfrac{6\times6}{6+6}=3\,\mu F\)
  3. \(Q=3\times12=36\,\mu C\)
  4. \(V_{block}=\dfrac{36}{6}=6\,V = V_{2\mu F}\)

Easy Trick

Two equal blocks in series (6 µF and \(2\!\parallel\!4=6\,\mu F\)) split the 12 V evenly → 6 V each. Every capacitor inside the parallel block sees that same 6 V, so the 2 µF reads 6 V directly.

b
\(V_{2\,\mu F} = 6\ V\)