Basic Electrical Engineering (3110005)

Question-2c-OR

BE | Semester-1 Winter-2019 | 11-01-2020

Q2) (c)

Charging of a Capacitor:

Initially, there is no charge on the capacitor. When the switch S is connected to terminal ‘a’ in Fig. 1, the capacitor gets charged

gradually to a potential of

V

volts.

Let

t

seconds time elapse. Then,

v

= instantaneous voltage across the capacitor in volts,

q

= charge on the capacitor plates in coulombs

i

= charging current in amperes,

According to KVL,

V = v + i R

\Rightarrow V = v + \frac{dq}{dt} R

\Rightarrow V = v + \frac{d (C v)}{dt} R

\Rightarrow V = v + C \frac{d v}{dt} R

\Rightarrow V - v = RC \frac{d v}{dt}

\Rightarrow \frac{d v}{V - v} = \frac{dt}{RC}

Multiplying both sides by -1,

\Rightarrow \frac{- d v}{V - v} = \frac{- dt}{RC}

Integrating both the sides,

\Rightarrow \int \frac{- d v}{V - v} = \int \frac{- dt}{RC}

\Rightarrow \int \frac{- d v}{V - v} = \frac{- 1}{RC} \int dt

\Rightarrow \log (V - v) = \frac{- t}{RC} + K

, where K = Constant

We know that initially when

t = 0, v = 0

\Rightarrow logV = 0 + K = K

Hence,

\Rightarrow \log (V - v) = \frac{- t}{RC} + logV

\Rightarrow \log (V - v) - logV = \frac{- t}{RC}

\Rightarrow \log (\frac{V - v}{V}) = \frac{- t}{RC}

\Rightarrow \frac{V - v}{V} = e^{\frac{- t}{RC}}

\Rightarrow V - v = V \times e^{\frac{- t}{RC}}

\Rightarrow v = V (1 - e^{\frac{- t}{RC}})

RC = λ

,then

\Rightarrow v = V (1 - e^{\frac{- t}{λ}})