An RC circuit consists of a resistance R = 5 MΩ and a capacitance C = 1.0 μF connected in series with a battery. In how much time will the potential diffrence across the capacitor become 8 times that across the resistor? (Given $l o g_{e} (3) = 1.1$ )
(IIT-JEE 2005)

5.5 s

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11 s

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44 s

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88 s

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Solution

The correct option is B
11 s

If the circuit is like this,

If $V_{c}$ is the potential differnce across C, the power differences across R will be $V - V_{C} .$
We are asked when will be
$8 (V - V_{c}) = V_{c}$
or, when is
$8 V = 9 V_{c}$ ...(1)
We know, for a capacitor
$q_{c} = q_{0} (1 - e^{- t / R C})$
$\frac{q_{c}}{C} = \frac{q_{0}}{C} (1 - e^{t / R C})$
$V_{c} = V (1 - e^{- t / R C}$
(since $\frac{q_{0}}{C} = V)$
$∴$ substitution in (1) gives,
$8 V = 9 V (1 - e^{- t / R C})$
$\frac{8}{9} = 1 - e^{- t / R C}$
$e^{- t / R C} = \frac{1}{9}$
$- t / R C = - l o g_{e} 9$
$t / R C = 2 l o g_{e} 3$
$t = 2 \times 1.1 \times 10^{- 6} \times 5 \times 10^{6}$
= 11 s

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An RC circuit consists of a resistance R = 5 MΩ and a capacitance C = 1.0 μF connected in series with a battery. In how much time will the potential diffrence across the capacitor become 8 times that across the resistor? (Given $l o g_{e} (3) = 1.1$ )
(IIT-JEE 2005)

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An RC circuit consists of a resistance R = 5 MΩ and a capacitance C = 1.0 μF connected in series with a battery. In how much time will the potential diffrence across the capacitor become 8 times that across the resistor? (Given loge(3)=1.1 ) (IIT-JEE 2005)

An RC circuit consists of a resistance R = 5 MΩ and a capacitance C = 1.0 μF connected in series with a battery. In how much time will the potential diffrence across the capacitor become 8 times that across the resistor? (Given $l o g_{e} (3) = 1.1$ )
(IIT-JEE 2005)