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Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t2 is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio $ \large\frac{t1}{t2}$ will be

$\begin {array} {1 1} (A)\;1 & \quad (B)\;\large\frac{1}{2} \\ (C)\;\large\frac{1}{4} & \quad (D)\;2 \end {array}$


1 Answer

$ q=q_0e^{\large\frac{-t}{t}}$
$U = \large\frac{q^2}{2C}$
$ \Rightarrow U = U_i \: e^{\large\frac{-2t}{t}}$
On substituting, $ q=\large\frac{q_0}{4}$ and $ U = \large\frac{U_i}{2}$, one gets the answer
Ans : (C)
answered Feb 13, 2014 by thanvigandhi_1
edited Mar 14, 2014 by thanvigandhi_1

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