Given, the peak voltage of the battery, V0 = 20V The voltage of the battery, V = 2V Time, t=1μs=1×10−6s Resistance of the capacitor, R = 10 Ω As we know that, V=V0(1−e−t∕RC) Substituting the values in the above equation, we get 2=20(1−e−t∕RC) ⇒
t
RC
=ln(
10
9
) ⇒ C=
t
Rln(
10
9
)
=
10−6
10×ln(
10
9
)
=0.95μF ∴ The capacitance of the capacitor is 0.95 μ F.