# Solved Problems-16

Problems-16

A 50Î¼F capacitor and 20000Î© resistor are connected in series across a 100V battery at

*t*= 0. At*t*= 0.5s, the battery voltage is suddenly increased to 150V. Find the charge on capacitor at*t*= 0.75s.Solution

When the circuit is connected to 100V supply, the equation of voltage across the capacitor is,

At

*t*= 0.5s, the voltage across the capacitor is,âˆ´ At

*t*= 0.5s, charge on the capacitor is,*q*=*Cv*= 50 Ã— 10_{C}^{â€“6}Ã— 39.347 = 1967.35 Ã— 10^{â€“6}CThis charge is the initial charge

*q*_{0}when the battery voltage is suddenly increased to 150V.When the circuit is connected to 150V, the KVL equation becomes,

Taking Laplace transform,

â‡’

Taking inverse Laplace transform,

Therefore, the voltage across the capacitor,

Substituting the values, the voltage across the capacitor at

*t*= 0.75s i.e., 0.25 second after changing the battery voltage,âˆ´ charge on the capacitor is,