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 red - Physics MCQ AIEEE 2026 | ExamTest.live | ExamTest.live

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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:

A $1$
B $1/2$
C $1/4$
D $2$

Solution & Step-by-step Explanation

Charge on a discharging capacitor: q = q_{0} e^{- t / RC} .Energy stored: U = \frac{q ^{2}}{2 C} = \frac{q _{0}^{2} e ^{- 2 t / RC}}{2 C} = U_{0} e^{- 2 t / RC} .Case 1: U = U_{0} /2 $ U_{0} /2 = U_{0} e^{- 2 t_{1} / RC} ⟹ 1/2 = e^{- 2 t_{1} / RC} $ ln (2) = \frac{2 t _{1}}{RC} ⟹ t_{1} = \frac{RC ln 2}{2} Case 2: q = q_{0} /4 $ q_{0} /4 = q_{0} e^{- t_{2} / RC} ⟹ 1/4 = e^{- t_{2} / RC} $ ln (4) = \frac{t _{2}}{RC} ⟹ 2 ln 2 = \frac{t _{2}}{RC} ⟹ t_{2} = 2 RC ln 2 Ratio: \frac{t _{1}}{t _{2}} = \frac{( RC ln 2 ) /2}{2 RC ln 2} = \frac{1}{4}

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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:

A

1

B

1/2

C

1/4

D

2

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