First slide

Capacitance

Question

In the network shown charge on the $2 μF$ capacitor is

Moderate

A

$52 μC$
B

$36 μC$
C

$27 μC$
D

$18 μC$

Solution

$x = (\frac{5}{2 + 3}) A = 1 A y = \frac{24}{(5 + 3)} A = 3 A ∴ V_{A} + 3 \times 1 - 14 - 5 \times 3 = V_{B} \Rightarrow V_{A} - V_{B} = 26 volt ∴ Q = 26 \times 2 μC = 52 μC$

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