Explanation Given data: Consider that no energy is stored in the circuit elements when the switch is closed. C = 100 nF The voltage across the capacitor is initially constant at 100 V. V = 100 V At t 1 = π 7 ms , the voltage across the capacitor is 163.84 V. v C ( t 1 ) = 163.84 V At t 3 = 3 π 7 ms , the voltage across the capacitor is 126.02 V. v C ( t 3 ) = 126.02 V Formula used: Write the expression of v C ( t ) for underdamped response in the given circuit as follows: v C ( t ) = V f + B ′ 1 e − α t cos ( ω d t ) + B ′ 2 e − α t sin ( ω d t ) , t ≥ 0 (1) Here, V f is the final value of the voltage, B ′ 1 , B ′ 2 are arbitrary constants, α is the neper frequency, and ω d is the damped frequency. Write the expression for damped frequency as follows: ω d = ( ω 0 ) 2 − α 2 (2) Write the expression for resonant radian frequency for the given circuit as follows: ω 0 = 1 L C (3) Here, L is the inductance of the inductor, and C is the capacitance of the capacitor. Write the expression for neper frequency for the given circuit as follows: α = R 2 L (4) Calculation: As the initial energy stored in the circuit is zero, the initial voltage across the capacitor in the circuit is zero. v C ( 0 ) = 0 (5) Differentiate on both sides of the expression to obtain the value of d v C ( 0 ) d t . d v C ( 0 ) d t = d d t ( 0 ) d v C ( 0 ) d t = 0 (6) At t < ∞ (steady state condition), the capacitor in the circuit gets open circuit. Therefore, the total source voltage appears across the capacitor. v C ( ∞ ) = V f v C ( ∞ ) = V V f = V (7) Substitute V for V f in Equation (1) as follows: v C ( t ) = V + B ′ 1 e − α t cos ( ω d t ) + B ′ 2 e − α t sin ( ω d t ) , t ≥ 0 (8) Substitute 0 for t to obtain the value of v C ( 0 ) . v C ( 0 ) = V + B ′ 1 e − α ( 0 ) cos [ ω d ( 0 ) ] + B ′ 2 e − α ( 0 ) sin [ ω d ( 0 ) ] = V + B ′ 1 ( 1 ) ( 1 ) + B ′ 2 ( 1 ) ( 0 ) = V + B ′ 1 From Equation (5), substitute 0 for v C ( 0 ) as follows: 0 = V + B ′ 1 Simplify the expression as follows: B ′ 1 = − V (9) Differentiate on both sides of the expression in Equation (8) as follows: d v C ( t ) d t = d d t [ V + B ′ 1 e − α t cos ( ω d t ) + B ′ 2 e − α t sin ( ω d t ) ] = { 0 − B ′ 1 e − α t [ α cos ( ω d t ) + ω d sin ( ω d t ) ] + B ′ 2 e − α t [ − α sin ( ω d t ) + ω d cos ( ω d t ) ] } = [ − B ′ 1 e − α t α cos ( ω d t ) − B ′ 1 e − α t ω d sin ( ω d t ) − α B ′ 2 e − α t sin ( ω d t ) + ω d B ′ 2 e − α t cos ( ω d t ) ] d v C ( t ) d t = e − α t [ ( ω d B ′ 2 − α B ′ 1 ) cos ( ω d t ) − ( ω d B ′ 1 + α B ′ 2 ) sin ( ω d t ) ] Substitute 0 for t to obtain the value of d v C ( 0 ) d t . d v C ( 0 ) d t = e − α ( 0 ) { ( ω d B ′ 2 − α B ′ 1 ) cos [ ω d ( 0 ) ] − ( ω d B ′ 1 + α B ′ 2 ) sin [ ω d ( 0 ) ] } = ( 1 ) [ ( ω d B ′ 2 − α B ′ 1 ) ( 1 ) − ( ω d B ′ 1 + α B ′ 2 ) ( 0 ) ] = ω d B ′ 2 − α B ′ 1 From Equation (6), substitute 0 for d i ( 0 ) d t as follows: 0 = − α B ′ 1 + ω d B ′ 2 B ′ 2 = α ω d B ′ 1 From Equation (9), substitute ( − V ) for B ′ 1 as follows: B ′ 2 = α ω d ( − V ) = − α V ω d Substitute t n for t in Equation (8) as follows: v C ( t n ) = V + B ′ 1 e − α t n cos ( ω d t n ) + B ′ 2 e − α t n sin ( ω d t n ) Consider t n = n π ω d and substitute n π ω d for t n on the right side of the expression as follows: v C ( t n ) = V + B ′ 1 e − α n π ω d cos ( ω d n π ω d ) + B ′ 2 e − α n π ω d sin ( ω d n π ω d ) = V + B ′ 1 e − α n π ω d cos ( n π ) + B ′ 2 e − α n π ω d sin ( n π ) = V + B ′ 1 e − α n π ω d ( − 1 ) n + B ′ 2 e − α n π ω d ( 0 ) { ∵ cos ( n π ) = ( − 1 ) n ∵ sin ( n π ) = 0 } = V + B ′ 1 e − α n π ω d ( − 1 ) n From Equation (9), substitute ( − V ) for B ′ 1 as follows: v C ( t n ) = V + ( − V ) e − α n π ω d ( − 1 ) n v C ( t n ) = V − V ( − 1 ) n e − α n π ω d (10) Substitute 1 for n to obtain the value of v C ( t 1 )