Question

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?

Answer 1

Explanation:

Provided,

Capacitor’s capacitance, C = 12 $\times$ 10−6 F

Battery’s Emf, V0 = 6.00 V

Battery’s internal resistance, R = 1 Ω

Time interval, t = 12 μs

(a)  In the circuit, the charging current is provided as,

i = i0e−t/RC

Current at, t = 12.0 μs

i = V0Re-t/RCi = 61 $\times$ e-1i = 2.21 A

(b)  The capacitor’s charge while charging at any time “t” is provided as

Q = CV0(1-e-tRC)  

Battery’s work in time delivering this charge,

W = QV0

By the battery, power delivered in time “t” is,

P = CV02(1-e-tRC)t

Putting, t = 12 μs

⇒P=13.25 W

(c)  In the capacitor, energy stored at any time is,

U = 12Q2C ⇒U = 12CV02(1-e-tRC)2

In the capacitor, the rate at which the energy stored is,

dUdt = 12CV02 $\times$ (e-tRC) $\times$ 2(1-e-tRC) $\times$ 1RC

⇒dUdt = V02R(e-tRC-e-2tRC)  

⇒ dUdt = 8.37 W

So, the dissipated power in heat = P - dUdt = 13.25 -8.37 = 4.87 W

(d) In the capacitor, the rate at which the energy deposited is increasing

⇒dUdt = 8.37 W