Concept:

The transient equation for the voltage across a capacitor is given by:

\({V_c}\left( t \right) = V\left( \infty \right) + \left[ {V\left( 0 \right) - V\left( \infty \right)} \right]{e^{ - \frac{t}{\tau }}}\)  ---(1)

V(∞) is the final voltage

V(0) is the initial voltage

τ is the time constant

Calculation:

The circuit diagram is as shown below.

Let the initial voltage across the capacitor is V₀₁

V(0) = V₀₁

The final voltage across the capacitor is 10 V, i.e.

V (∞) = 10 V

Using Equation (1), we get:

\({V_c}\left( t \right) = 10 + \left[ {{V_{01}} - 10} \right]{e^{ - \frac{t}{\tau }}}\)

At t = 0.4τ, voltage across capacitor is zero i.e. V_C(t)=0

\( \Rightarrow 0 = 10 + \left[ {{V_{01}} - 10} \right]{e^{ - \frac{{0.4\tau }}{\tau }}}\)

\( \Rightarrow {V_{01}} = 10 - 10/{e^{-0.4}} = - 4.918\;V\)

At t = 0.2τ, for the voltage across the capacitor to become zero i.e. For V_C(t) = 0, at t = 0.2τ, we assume that the initial voltage is V­₀₂

Using the same Equation, we can write:

\( \Rightarrow 0 = 10 + \left[ {{V_{02}} - 10} \right]{e^{ - \frac{{0.2\tau }}{\tau }}}\)

\( \Rightarrow {V_{02}} = 10 - 10/{e^{-0.2}} = - 2.214\;V\)

The percentage change required in the initial capacitor voltage will be:

\(\frac{{{V_{02}} - {V_{01}}}}{{{V_{01}}}} \times 100\)

\( = \frac{{ - 2.214 + 4.918}}{{4.918}} \times 100\)

= 54.98 %

The absolute value of % change is 54.98 %.