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AIEEE 2010 Solved Paper

Section: Physics

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Question : 43 of 89

Marks: +1, -0

Let

C

be the capacitance of a capacitor discharging through a resistor

R

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

\frac{t_1}{t_2}

$1$
$\frac{1}{2}$
$\frac{1}{4}$
$2$

Solution:

Initial energy of capacitor,

E_1 = \frac{q_1^2}{2C}

Final energy of capacitor,

E_2 = \frac{1}{2} E_1 = \frac{q_1^2}{4C} = \left( \frac{\frac{q_1}{\sqrt{2}}}{2C} \right)^2

\therefore t_1 =

time for the charge to reduce to

\frac{1}{\sqrt{2}}

of its initial value

and

t_2 =

time for the charge to reduce to

\frac{1}{4}

of its initial value

We have,

q_2 = q_1 e^{-\frac{t}{C R}}

\Rightarrow \ln \left( \frac{q_2}{q_1} \right) = -\frac{t}{C R}

\therefore \ln \left( \frac{1}{\sqrt{2}} \right) = \frac{-t_1}{C R} \dots (1)

and

\ln \left( \frac{1}{4} \right) = \frac{-t_2}{C R} \dots (2)

By (1) and

(2), \frac{t_1}{t_2} = \frac{\ln \left( \frac{1}{\sqrt{2}} \right)}{\ln \left( \frac{1}{4} \right)}

= \frac{1}{2} \frac{\ln \left( \frac{1}{2} \right)}{2 \ln \left( \frac{1}{2} \right)} = \frac{1}{4}

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