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Question
A capacitor of 4 µ F is connected as shown in the circuit (Figure). The internal resistance of the battery is 0.5 Ω. The amount of charge on the capacitor plates will be ______.
Options
0
4 µ C
16 µ C
8 µ C
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Solution
A capacitor of 4 µ F is connected as shown in the circuit (Figure). The internal resistance of the battery is 0.5 Ω. The amount of charge on the capacitor plates will be 8 µ C.
Explanation:
A capacitor offers zero resistance in a circuit when it is uncharged, i.e., it can be assumed as short-circuited and it offers infinite resistance when it is fully charged.
At steady state
At steady state the capacitor offers infinite resistance in the DC circuit and acts as the open circuit as shown in the figure, therefore no current flows through the capacitor and 10 Ω resistance, leaving zero potential difference across 10 Ω resistance. Hence potential difference across capacitors will be the potential difference across A and B.
The potential difference across lower and middle branches of the circuit is equal to the potential difference across the capacitor of upper branch of circuit.
Current flows through 2 Ω resistance from left to right, is given by I = v/R + r = 1A.The potential difference across 2 Ω resistance, V = IR = 1 × 2 = 2V Hence potential difference across capacitors is also 2V.
The charge on capacitor is q = CV = (2 μ F) × 2V = 8 μ C.
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