Question

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 t1/t2 will be

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{6}$

Answer 1

Answer: (D)

$\frac{1}{6}$

Answer 2

The correct answer is (D) : $\frac{1}{6}$
For a discharging capacitor when energy reduces to half the charge would become 1/√2 times the initial value.
$⇒ ( \frac{1}{2} )^{1/2} = e^{-t1/r}$
Similarly,
$( \frac{1}{2} )^3 = e^{-t2/r}$
$⇒ \frac{t1}{t2} = \frac{1}{6}$