JEE Main · 2010 · 11 Apr 2010 · Offline · Current Electricity

Let C be the capacitance of a capacitor discharging through a resistor R. Suppose

t_{1}

is the time taken for the energy stored in the capacitor to reduce to half its initial value and

t_{2}

is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio

t_{1} / t_{2}

will be

Answer: (C)

Solution

When a charged capacitor discharges through a resistor, the charge decays exponentially as

q = q_{0} e^{- t / τ}

with time constant

τ = R C

. The energy stored is

U = q^{2} /2 C

, so it falls as

U = U_{0} e^{- 2 t / τ}

, twice as fast as the charge because energy goes as the square of charge. For the energy to drop to half we set

e^{- 2 t_{1} / τ} = \frac{1}{2}

, giving

t_{1} = \frac{τ}{2} ln 2

. For the charge to fall to one quarter we set

e^{- t_{2} / τ} = \frac{1}{4}

, giving

t_{2} = τ ln 4 = 2 τ ln 2

. Dividing, the time constant and the

ln 2

cancel, leaving

t_{1} / t_{2} = \frac{1}{4}

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option