First slide

Capacitance

Question

Two capacitors 2 $μ$ F and 4 $μ$ F are connected in parallel. A third capacitor of 6 $μ$ F capacity is connected in series. The combination is then connected across a 12 V battery. The voltage across 2 $μ$ F capacity is

Moderate

A

2 V
B

6 V
C

8 V
D

1 V

Solution

Resultant capacitance of condensers of capacity 2 $μ$ F and 4 $μ$ F when connected in parallel $C^{'} = 2 + 4 = 6 μF$
This is connected in series with a capacitor of capacity 6 $μ$ F in series. The resultant capacity C ls given by
$\frac{1}{C} = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} or C = 3 μF$
Charge on combination
$q = (3 \times 10^{- 6}) \times (12) = 36 \times 10^{- 6} C$
Let the charge on 2 $μ$ F capacitor be q₁ then
$\begin{array}{l} \frac{q_{1}}{2} = \frac{q - q_{1}}{4} or q_{1} = \frac{q}{3} \\ ∴ q_{1} = 12 \times 10^{- 6} C \end{array}$
Now potential across 2 $μ$ F condenser
$= \frac{q_{1}}{2 \times 10^{- 6}} = \frac{12 \times 10^{- 6}}{2 \times 10^{- 6}} = 6 V$

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