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 in the capacitor reduces to half of its initial value and in time t2 the charge stored reduces to one eight of its initial value. The ratio t1/t2 will be

According to Law of discharging of capacitor, we can write Q = Q 0 e − t τ ⇒ Q 2 8 = Q 0 e − t 2 τ ⇒ t 2 3 τ l n ( 2 ) , a n d U = Q 2 2 C = Q 0 2 2 C e − 2 t τ ⇒ 1 2 ( Q 0 2 2 C ) = Q 0 2 2 C e − 2 t τ ⇒ t 1 = τ 2 l n ( 2 ) ⇒ t 1 t 2 = 1 6

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Electrostatics Physics Electrostatic Potential and Capacitance Physics Class 12th

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 in the capacitor reduces to half of its initial value and in time t₂ the charge stored reduces to one eight of its initial value. The ratio t₁/t₂ will be

Option 1 -

1/2

Option 2 -

1/3

Option 3 -

1/4

Option 4 -

1/6

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 weeks ago

Correct Option - 4

Detailed Solution:

According to Law of discharging of capacitor, we can write

$Q = Q_{0} e^{- \frac{t}{τ}} \Rightarrow \frac{Q_{2}}{8} = Q_{0} e^{- \frac{t_{2}}{τ}} \Rightarrow t_{2} 3 τ l n (2), a n d$

$U = \frac{Q^{2}}{2 C} = \frac{Q_{0}^{2}}{2 C} e^{- \frac{2 t}{τ}} \Rightarrow \frac{1}{2} (\frac{Q_{0}^{2}}{2 C}) = \frac{Q_{0}^{2}}{2 C} e^{- \frac{2 t}{τ}} \Rightarrow t_{1} = \frac{τ}{2} l n (2)$

$\Rightarrow \frac{t_{1}}{t_{2}} = \frac{1}{6}$

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