A capacitor of capacitance 12.0 μF is connected to a battery of emf 6.00 V and internal resistance 1.00 Ω through resistanceless leads. 12.0 μs after the connections are made, what will be (a) the current in the circuit (b) the power delivered by the battery (c) the power dissipated in heat and (d) the rate at which the energy stored in the capacitor is increasing?

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Solution

Given,
Capacitance of capacitor, C= 12.0 μF = 12 × 10⁻⁶ F
Emf of battery, V₀ = 6.00 V
Internal resistance of battery, R = 1 Ω
Time interval, t = 12 μs

(a)
Charging current in the circuit is given as,
i = i₀e^−t^/RC
Current at, t = 12.0 μs
$i = \frac{V_{0}}{R} e^{- t / R C} i = \frac{6}{1} \times e^{- 1} i = 2.207 = 2.21 A$

(b)
During charging, charge on the capacitor at any time ''t'' is given as
$Q = C V_{0} (1 - e^{- \frac{t}{R C}})$
Work done by battery in in time delivering this charge is,
W = QV₀

Power deliver by the battery in time ''t'' is,
$P = \frac{C {V_{0}}^{2} (1 - e^{- \frac{t}{R C}})}{t}$
Putting, t = 12 μs
$P = \frac{12 \times 10^{- 6} {V_{0}}^{2} (1 - e^{- 1})}{12 \times 10^{- 6}} \Rightarrow P = 13.25 W$

(c)
Energy stroed in the capacitor at any instant of time is given as,
$U = \frac{1}{2} \frac{Q^{2}}{C} \Rightarrow U = \frac{1}{2} \frac{C^{2} {V_{0}}^{2} (1 - e^{- \frac{t}{R C}})^{2}}{C} \Rightarrow U = \frac{1}{2} C {V_{0}}^{2} (1 - e^{- \frac{t}{R C}})^{2}$

Rate at which the energy stored in the capacitor is,
$\frac{d U}{d t} = \frac{1}{2} C {V_{0}}^{2} \times 2 (1 - e^{- \frac{t}{R C}}) \times (e^{- \frac{t}{R C}}) \times \frac{1}{R C}$
$\Rightarrow \frac{d U}{d t} = \frac{{V_{0}}^{2}}{R} (e^{- \frac{t}{R C}} - e^{- \frac{2 t}{R C}}) \Rightarrow \frac{d U}{d t} = \frac{6 \times 6}{1} (e^{- 1} - e^{- 2}) \Rightarrow \frac{d U}{d t} = 8.37 W$

So, the power dissipated in heat = $P - \frac{d U}{d t}$
= 13.25 $-$ 8.37 = 4.87 W

(d)
Rate at which the energy stored in the capacitor is increasing = $\Rightarrow \frac{d U}{d t} = 8.37 W$

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