First slide

Capacitance

Question

In the network shown, if the potential difference between the plates of $3 μF$ capacitor is 5 volt, the charge stored in $2 μF$ capacitor is

Moderate

A

22.5 $μC$
B

10 $μC$
C

7.5 $μC$
D

15 $μC$

Solution

Charge on $3 μF$ capacitor $= Q_{1} = 3 \times 5 μ C = 15 μ C$
$∴$ Charge on $3 μF$ capacitor $= 15 μC$
$C_{e} =$ Equivalent capacitance of $3 μF$ and $6 μF$ capacitors
$= \frac{3 \times 6}{3 + 6} μF = 2 μF$
$∴ V_{A} - V_{B} = \frac{Q}{C_{e}} = \frac{15}{2} volt =7.5 volt$
$∴$ Charge on $2 μF$ capacitor $= 2 \times (V_{A} - V_{B}) = 2 \times 7 .5 μC = 15 μC$

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