Question

The charge accumulated on the capacitor connected in the following circuit is ____ µC
(Given $C = 150 \, \mu \text{F}$)

Answer 1

Using Kirchhoff’s Voltage Law (KVL) in the circuit:

$V_A + \frac{10}{3}(1) - 6(1) = V_B$

Thus,

$V_A - V_B = 6 - \frac{10}{3} = \frac{8}{3} \text{V}$

The charge $Q$ on the capacitor is:

$Q = C(V_A - V_B) = 150 \times \frac{8}{3} = 400 \, \mu C$