A constant current of 2 mA charges a 20 μF capacitor for 2s. Which of the following is true for the charging of the capacitor.

Concept:

The current across a capacitor is defined as:

\(i = C.\frac{{d{V_c}}}{{dt}}\)

Rearranging the above, we can write:

\(\frac{{d{V_c}}}{{dt}} = \frac{i}{C}\)

Integrating both sides we get:

\({V_c}\left( t \right) = \frac{1}{C}\mathop \smallint \nolimits_0^t i.dt\)

Calculation:

Given i = 2 mA (Constant)

C = 20 μF

The capacitor voltage will be given by:

\({V_c}\left( t \right) = \frac{1}{20~\mu F}\mathop \smallint \nolimits_0^t 2~mA~dt\)

\({V_c}\left( t \right) = \frac{1}{20~\mu F}.2~mA(t)|_0^t\)

\({V_c}\left( t \right) =100t\)

This indicates a linear variation of the voltage across the capacitor.

The voltage after t = 2 sec will be:

\({V_c}\left( 2 \right) =100\times 2=200~V\)

So, the capacitor voltage increases linearly from 0 to 200 V in 2 sec.