In an electrical circuit drawn below the amount of charge stored in the capacitor is _______ μC .

Question

Aicharya · Accepted Answer

In steady state there will be no current in branch of capacitor, so no voltage drop across $R_{2} = 5 Ω$

$I_{2} = 0$

$I_{1} = I_{3} = \frac{10}{4 + 6} = 1 A$

$V_{R_{3}} = V_{c} + V_{R_{2}} (V_{R_{2}} = 0)$

$\Rightarrow I_{3} R_{3} = V_{c}$

$\Rightarrow V_{c} = 1 \times 6 = 6$ volt

Therefore, charge on capacitor $q_{c} = C V_{c} = 10 \times 6 = 60 μC$

Solution