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Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorCircuit AnalysisMedium3 minQB

PhysicsMediumsingle choice

In the circuit shown, $C_{1} = 6 μ$ F, $C_{2} = 3 μ$ F and battery $B = 20$ V. The switch $S_{1}$ is first closed. It is then opened and afterwards $S_{2}$ is closed. What is the charge on $C_{2}$ ?

Options:

Answer:

Solution:

Initially:
$C_{1} = 6 \times 1 0^{- 6} \times 20 = 120 μ$ C
Finally:
$V = \frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}} = \frac{120 \times 1 0 ^{- 6}}{9 \times 1 0 ^{- 6}} = \frac{40}{3}$ V
Charge on $C_{2}$ :
$C_{2} = \frac{40}{3} \times 3 μ$ F $= 40 μ$ C

Subject:PhysicsTopic:CapacitorSubtopic:Circuit Analysis

⏱ 3mℹ️ Source: QB

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