First slide

KVL

Question

In the circuit shown in the figure, if switches S₁ and S₂ have been closed for a long time, then charge on the capacitor

Moderate

A

1s $100 μ C$
B

increases to $200 μ C$ if one third of the gap of the capacitor's plates is filled with a dielectric (K = 2) of same area
C

remains unchanged if one third of the gap of the capacitor's plates is filled with,a dielectric (K = 2) of same area.
D

both (1) and (2)

Solution

After long time, current through the capacitor = 0
Current through $6 Ω resistor, i = \frac{12 - 4}{8} = 1 A$
Voltage across capacitor, V = 4 + (6) (1) = 10 V
Charge on the capacitor, $Q = (10) (10) = 100 μC$
After the insertion of dielectric
$\begin{array}{l} C^{'} = \frac{{Aε}_{0}}{\frac{d}{3} + \frac{d}{3} + \frac{d}{3 (2)}} = \frac{{Aε}_{0}}{d} [\frac{1}{(\frac{1}{3} + \frac{1}{3} + \frac{1}{6})}] \\ = (10) (\frac{1}{(5 / 6)}) = \frac{6}{5} (10) = 12 μF \end{array}$
Here voltage across capacitor remains same.
$∴ Charge on the capacitor, Q' = (12) (10) = 120 μC$

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