A capacitor is discharging through a resistor R . Consider in time t 1 , the energy stored in the capacitor reduces to half of its initial value and in time t 2 , the charge stored reduces to one eighth of its initial value. The ratio t 1 t

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A capacitor is discharging through a resistor R . Consider in time t 1 , the energy stored in the capacitor reduces to half of its initial value and in time t 2 , the charge stored reduces to one eighth of its initial value. The ratio t 1 t 2 will be

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Correct answer: D. 1 6

We know that in the discharging circuit the charge is given by, q = q 0 e - t R C &#8658; t R C = ln q 0 q In time t 1 the energy stored reduces to half. Hence, q 1 2 2 C = 1 2 &#215; q 0 2 2 C &#8658; q 0 q 1 = 2 . Therefore, t 1 R C = ln 2 = 1 2 ln 2 . In time t 2 the charge stored reduces to 1 8 th of initial value. Hence, q 2 q 0 = 1 8 . Therefore, t 2 R C = ln 8 = 3 ln 2 . Hence, t 1 t 2 = 1 2 ln 2 3 ln 2 = 1 6

A capacitor is discharging through a resistor R . Consider in time t 1 , the energy stored in the capacitor reduces to half of its initial value and in time t 2 , the charge stored reduces to one eighth of its initial value. The ratio t 1 t 2 will be

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