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JEE Main 2010
Exam
JEE Main
Year
2010
Subject
Physics
Chapter
Capacitance

JEE Main 2010 · Physics · Capacitance

Let C be the capacitance of a capacitor discharging through a...

Let CC be the capacitance of a capacitor discharging through a resistor R. Suppose t1t_1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t2t_2 is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio t1/t2t_1 / t_2 will be

Options

  1. A1
  2. B12\frac{1}{2}
  3. C14\frac{1}{4}
  4. D2

Correct answer

C. 14\frac{1}{4}

Step-by-step solution

U=12q2C=12C(q0e−t/T)2=q022Ce−2t/T(\mathrm{U}=\frac{1}{2} \frac{\mathrm{q}^2}{\mathrm{C}}=\frac{1}{2 \mathrm{C}}\left(\mathrm{q}_0 \mathrm{e}^{-\mathrm{t} / \mathrm{T}}\right)^2=\frac{\mathrm{q}_0^2}{2 \mathrm{C}} \mathrm{e}^{-2 \mathrm{t} / \mathrm{T}} \quad( where τ=CR)\tau=\mathrm{CR}) U=Uie−2t/τ\mathrm{U}=\mathrm{U}_{\mathrm{i}} \mathrm{e}^{-2 \mathrm{t} / \mathrm{\tau}} 12Ui=Uie−2t1/τ\frac{1}{2} \mathrm{U}_{\mathrm{i}}=\mathrm{U}_{\mathrm{i}} \mathrm{e}^{-2 \mathrm{t}_1 / \mathrm{\tau}} 12=e−2t1/τ⇒t1=T2ln⁡2\frac{1}{2}=\mathrm{e}^{-2 \mathrm{t}_1 / \tau} \Rightarrow \mathrm{t}_1=\frac{\mathrm{T}}{2} \ln 2 Now \quad\mathrm{q}=\mathrm{q}0e−t/T_0 \mathrm{e}^{-\mathrm{t} / \mathrm{T}} 14q0=q0e−t/2 T\frac{1}{4} \mathrm{q}_0=\mathrm{q}_0 \mathrm{e}^{-\mathrm{t} / 2 \mathrm{~T}} t2=T⁡ln⁡4=2T⁡ln⁡2\mathrm{t}_2=\operatorname{T} \ln 4=2 \operatorname{T} \ln 2 ∴t1t2=14\therefore \quad \frac{\mathrm{t}_1}{\mathrm{t}_2}=\frac{1}{4}
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