Name: RC Circuit
Uploaded: 2019-07-19
Duration: 4 min 3 s
Description: In this video, let us discuss how an RC circuit behaves when the voltage source is applied and also the terms and parameters involved in the analysis of it.

Question

A $3.00 MΩ$ resistor and a $1.00 μ F$ capacitor are connected in series with an ideal battery of emf $ε = 4.00 V$ . At $6.00 s$ after the connection is made, what is the rate at which

(a) the charge of the capacitor is increasing?

(b) energy is being stored in the capacitor?

(c) thermal energy is appearing in the resistor?

(d) energy is being delivered by the battery?

Accepted Answer

The variation of charge on the capacitor plate of capacitor,

$q = C ε (1 - e^{- \frac{t}{τ}})$

Where, C is the capacitance of capacitor,t,time constant $ε$ , emf of a battery.

The time constant $τ = R C$

Since, resistance, $R = 3 MΩ = 3 \times 10^{6} Ω$

Capacitance of capacitor, $C = 1 μF$ $= 1 \times 10^{- 6} F$

Time constant, $τ = (3 \times 10^{6} Ω) (1 \times 10^{- 6} F)$ $= 3 s$

$a t t = 6 s, \frac{t}{τ} = \frac{6}{3} = 2$

The rate of charge increasing

$\frac{d q}{d t} = \frac{d}{d t} (C ε (1 - e^{\frac{- t}{τ}}))$

$= - C ε e^{\frac{- t}{τ}} \times \frac{- 1}{τ}$

$= \frac{c ε}{τ} e^{\frac{- t}{τ}}$

Since $C = 1 \times 10^{- 6} F, ε = 4 V, τ = 3 s$

$\frac{d q}{d t} = \frac{(1 \times 10^{- 6} F) \times (4 V)}{3} e^{- 2}$ $= 1.7 \times 10^{- 7} A$

(b) The energy stored in the capacitor is given by, $U_{c} = \frac{q^{2}}{2 C}$

The rate of energy change

$\frac{d U_{c}}{d t} = \frac{d}{d t} (\frac{q^{2}}{2 C})$

$= \frac{q}{C} \frac{d q}{d t}$

But the charge, $q = C ε (1 - e^{\frac{- t}{τ}})$

$= (1 \times 10^{- 6} F) (4 V) (1 - e^{\frac{- 6}{3}})$

$= 3.48 \times 10^{- 6} C$

$\frac{d U_{c}}{d t} = \frac{q}{C} \frac{d q}{d t}$

$= (\frac{3.48 \times 10^{- 6} C}{1 \times 10^{- 6} F}) (1.7 \times 10^{- 7} A)$

$= 5.916 \times 10^{- 7} W$

Therefore, the energy delivered by the battery per second, $5.916 \times 10^{- 7} W$ .

(c) The rate at which energy is being dissipated in the resistor is given by $P = i^{2} R$ . The current is $1.7 \times 10^{- 7} A$ , so

$P = {(1.7 \times 10^{- 7} A)}^{2} \times (3 \times 10^{6} Ω) = 8.67 \times 10^{- 8} W$

(d) The rate at which energy is delivered by the battery is

$i ε = 1.7 \times 10^{- 7} A \times 4 V = 6.8 \times 10^{- 7} W$

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