JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Current Electricity

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 :

Answer: (D)

Solution

For a discharging capacitor:

q = Q_{0} e^{- t / R C}

Energy

U = \frac{q ^{2}}{2 C}

, so when energy becomes half, charge becomes

\frac{1}{2}

times.

At time

t_{1}

q = \frac{Q _{0}}{2} = Q_{0} e^{- t_{1} / R C}

This gives:

e^{- t_{1} / R C} = \frac{1}{2} = 2^{- 1/2}

At time

t_{2}

q = \frac{Q _{0}}{8} = Q_{0} e^{- t_{2} / R C}

This gives:

e^{- t_{2} / R C} = \frac{1}{8} = 2^{- 3} = (2^{- 1/2})^{6}

Therefore:

t_{2} = 6 t_{1}

, so

\frac{t _{1}}{t _{2}} = \frac{1}{6}

Answer: (D)

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Single correct option