Question

The voltage applied across a capacitor having a capacitance of 10μF is varied thus : The p.d. is increased uniformly from 0 to 600 V in seconds. It is then maintained constant at 600 V for 1 second and subsequently decreased uniformly to zero in five seconds. Plot a graph showing the variation of current during these 8 seconds. Calculate (a) the charge (b) the energy stored in the capacitor when the terminal voltage is 600.

Answer 1

The variation of voltage across the capacitor is as shown in Fig.. 

The charging current is given by

i = dq/dt = d/dt(CV) = C(dv/dt)

Charging current during the first stage 

= 10 × 10⁻⁶ × (600/2) = 3 × 10⁻³ A = 3 mA 

Charging current during the second stage is zero because dv/dt = 0 as the voltage remains constant. 

Charging current through the third stage

= 10 x 10^-6 x ((0 - 600)/5) = − 1.2 × 10⁻³ A = − 1.2 mA

The waveform of the charging current or capacitor current is shown in Fig..

 

(a) Charge when a steady voltage of 600 V is applied is 

= 600 × 10 × 10⁻⁶ = 6 × 10⁻³ C 

(b) Energy stored = 1/2 C V² = 1/2 × 10⁻⁵ × 600² = 1.8J