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…
JEE Main 2022 — Physics Electromagnetism
2022mcqmedium
A capacitor is discharging through a resistor R. Consider in time t1, the energy stored in the capacitor reduces to half of its initial value and in time t2, the charge stored reduces to one eighth of its initial value. The ratio t2t1 will be
Official previous-year question
Held on 29 Jun 2022 · Verified 6 Jul 2026.
Options
A
21
B
31
C
41
D
61
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Solution
We know that in the discharging circuit the charge is given by,
q=q0eRC−t⇒RCt=ln(qq0)
In time t1 the energy stored reduces to half. Hence, 2Cq12=21×2Cq02⇒q1q0=2. Therefore,
RCt1=ln(2)=21ln2.
In time t2 the charge stored reduces to 81th of initial value. Hence, q0q2=81. Therefore,
RCt2=ln(8)=3ln2.
Hence,
t2t1=3ln221ln2=61
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