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Question
A 100 μF capacitor is joined to a 24 V battery through a 1.0 MΩ resistor. Plot qualitative graphs (a) between current and time for the first 10 minutes and (b) between charge and time for the same period.
Solution
Time constant of the circuit,
τ = RC
= 1 × 106 × 100 × 10−6
= 100 s
The growth of charge through a capacitor,
\[q = VC \left(1 - e^{- \frac{t}{RC}}\right)\]
The current through the circuit,
\[i = \frac{dq}{dt}\]
\[ = \frac{VC}{RC} \cdot e^{- \frac{t}{RC}} \]
\[ = \frac{V}{R} . e^{- \frac{t}{RC}} \]
\[ = 24 \times {10}^{- 6} \cdot e^{- \frac{t}{100}}\]
For t = 10 min = 600 s
q = 24 × 10−4 × (1 − e−6)
= 23.99 × 10−4
\[i = \frac{24}{{10}^6} . \frac{1}{e} = 5 . 9 \times {10}^{- 8} A\]
(a) The plot between current and time for the first 10 minutes is shown below.
(b) The plot between charge and time for the first 10 minutes is shown below.
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