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Question
A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 360 μC. When potential across the capacitor is reduced by 120 V, the charge stored in it becomes 120 μC.
Calculate:
(i) The potential V and the unknown capacitance C.
(ii) What will be the charge stored in the capacitor, if the voltage applied had increased by 120 V?
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Solution
(i) Initial voltage, V1 = V volts and charge stored, Q1 = 360 µC.
Q1 = CV1 …(1)
Changed potential, V2 = V − 120
Q2 = 120 µC
Q2 = CV2 ...(2)
By dividing (2) from (1), we get `Q_1/Q_2 = (CV_1)/(CV_1) ⇒360/120 = V/(V-120)⇒ V= 180` volts
`∴C = Q_1/V_1 =( 360 xx10^-6)/180 = 2 xx 10^-6 F = 2μF`
(ii) If the voltage applied had increased by 120 V, then V3 = 180 + 120 = 300 V
Hence, charge stored in the capacitor, Q3 = CV3 = 2×10-6×300 = 600 µC
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