Question

Consider a capacitor-charging circuit. Let Q₁ be the charge given to the capacitor in a time interval of 10 ms and Q₂ be the charge given in the next time interval of 10 ms. Let 10 μC charge be deposited in time interval t₁ and another 10 μC charge be deposited in the next time interval t₂.

Options

• Q₁ > Q₂, t₁ > t₂

• Q₁ > Q₂, t₁ < t₂

• Q₁ < Q₂, t₁ > t₂

• Q₁ < Q₂, t₁ < t₂

Answer 1

Q₁ > Q₂, t₁ < t₂

 

The charge Q on the plates of a capacitor at time t after connecting it in a charging circuit,

\[Q = \epsilon C(1 - e^{- t/RC} )\]

where

ε = emf of the battery connected in the charging circuit

C = capacitance of the given capacitor

R = resistance of the resistor connected in series with the capacitor

The charge developed on the plates of the capacitor in first 10 mili seconds is given by

\[Q_1  = \epsilon C(1 -  e^{- 10 \times {10}^{- 3} /RC} )\]

The charge developed on the plates of the capacitor in first 20 mili seconds is given by

\[Q' = \epsilon C(1 - e^{- 20 \times {10}^{- 3} /RC} )\]

The charge developed at the plates of the capacitor in the interval t = 10 mili seconds to 20 mili seconds is given by

\[Q_2  = Q' -  Q_1 \]

\[ Q_2  = \left[ \epsilon C(1 - e^{- 20 \times {10}^{- 3} /RC} ) \right]   -   \left[ \epsilon C(1 - e^{- 10 \times {10}^{- 3} /RC} ) \right]\]

\[ \Rightarrow  Q_2  = \left[ \epsilon C( e^{- 10 \times {10}^{- 3} /RC} - e^{- 20 \times {10}^{- 3} /RC} ) \right]\]

\[ \Rightarrow  Q_2  = \left[ \epsilon C( e^{- 10 \times {10}^{- 3} /RC} )(1 - e^{- 10 \times {10}^{- 3} /RC} ) \right]\]

Comparing Q₁ with Q₂

\[Q_1 = \epsilon C(1 - e^{- 10 \times {10}^{- 3} /RC} ), Q_2 = \epsilon C e^{- 10 \times {10}^{- 3} /RC} (1 - e^{- 10 \times {10}^{- 3} /RC} )\]

\[\frac{Q_1}{Q_2} = \frac{1}{e^{- 10 \times {10}^{- 3} /RC}}\]

\[ e^{- 10 \times {10}^{- 3} /RC} < 1\]

\[ \therefore Q_1 > Q_2\]

For second part of the question

10 μC charge be deposited in a time interval t₁ and the next 10 μC charge is deposited in the next time interval t₂.
The time taken for 10 μC charge to develop on the plates of the capacitor is t₁

\[10  \mu C = \epsilon C(1 -  e^{- t_1 /RC} )............(1)\]

The time taken for 20 μC charge to develop on the plates of the capacitor is t'

\[20 \mu C = \epsilon C(1 - e^{- t_2 /RC} )............(2)\]

Dividing (2) by (1)

\[\frac{20  \mu C}{10  \mu C} = \frac{\epsilon C(1 - e^{- t_2 /RC} )}{\epsilon C(1 - e^{- t_1 /RC} )}\]

\[2 = \frac{(1 - e^{- t_2 /RC} )}{(1 - e^{- t_1 /RC} )}\]

\[2(1 -  e^{- t_1 /RC} )   =   (1 -  e^{- t_2 /RC} )\]

\[   e^{- t_2 /RC}  = 2 e^{- t_1 /RC}  - 1\]

Taking natural log both side,

\[\frac{t_2}{RC} = \ln(2) + \frac{t_1}{RC}\]

⇒ t₁ < t₂