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Question
A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 300 μC. When potential across the capacitor is reduced by 100 V, the charge stored in it becomes 100 μC. Calculate The potential V and the unknown capacitance. What will be the charge stored in the capacitor if the voltage applied had increased by 100 V?
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Solution
(i) Initial voltage, V1 = V volts, Charge stored,Q1 = 300 µC
Q1 = CV1 ..(1)
Changed potential, V2 = V-100 V
Q2 = 100 µC
Q2 = CV2 ..(2)
By dividing (2) from (1), we get `(Q_1)/(Q_2) = (CV_1)/(CV_2) => Q_1/Q_2 = V_1 /V_2`
`=> 300/100 = V/(V- 100) => V = 150 \text { volts }`
`∴ C = Q_1/V_1 = (300 xx 10^-6)/150 = 2 xx 10^-6 F = 2 μ F`
(ii) If the voltage applied had increased by 120 V, then V3 = 150+100 = 250 V
Hence, charge stored in the capacitor, Q3 = CV3 = 2×10-6×250 = 500 µC
(ii) If the voltage applied had increased by 120 V, then V3 = 150+100 = 250 V
Hence, charge stored in the capacitor, Q3 = CV3 = 2×10-6×250 = 500 µC
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