Explanation Given data: Refer to given circuit in the textbook. Now energy stored in the circuit at the instant the switch closes. Formula used: Write a general expression to calculate the impedance of a resistor in s -domain. Z R = R (1) Here, R is the value of resistance. Write a general expression to calculate the impedance of an inductor in s -domain. Z L = s L (2) Here, L is the value of inductance. Write a general expression to calculate the impedance of a capacitor in s -domain. Z C = 1 s C (3) Here, C is the value of capacitance. Calculation: The given circuit is redrawn as shown in Figure 1. Since no energy stored in the circuit at instant the switch closes, the current through inductor and voltage across the capacitor is equal to zero. i ( 0 ) = 0 A v ( 0 ) = 0 V For time t > 0 , the switch closes and the circuit in Figure 1 is reduced as shown in Figure 2. Use equation (1) to find Z R 1 . Z R 1 = 1 k Ω = 1 × 10 3 { ∵ 1 k = 10 3 } = 1000 Use equation (1) to find Z R 2 . Z R 2 = 1 k Ω = 1 × 10 3 { ∵ 1 k = 10 3 } = 1000 Substitute 10 H for L in equation (2) to find Z L . Z L = s ( 10 H ) = 10 s Substitute 2 μ F for C in equation (3) to find Z C . Z C = 1 s ( 2 μ F ) = 1 s ( 2 × 10 − 6 F ) { ∵ 1 μ = 10 − 6 } = 5 × 10 5 s Convert the Figure 2 into s -domain as shown in Figure 3. To obtain the Thevenin impedance short circuit the voltage source and remove the load capacitor in Figure 3. Now, the Figure 3 is reduced as shown in Figure 4. Refer to Figure 4, the impedance of resistor R 2 is connected in series with the parallel combination of impedance of resistor R 1 and inductor L . The Thevenin impedance Z Th is calculated as follows: Z Th = 1000 + ( 1000 ∥ 10 s ) = 1000 + 10000 s 1000 + 10 s = 1000 ( 1000 + 10 s ) + 10000 s 1000 + 10 s = 1000 ( 1000 + 10 s + 10 s ) 1000 + 10 s Simplify the above equation as follows: Z Th = 1000 ( 1000 + 20 s ) 1000 + 10 s = 20000 ( s + 50 ) 10 ( s + 100 ) = 2000 ( s + 50 ) s + 100 To find the Thevenin voltage open circuit the load capacitor in Figure 3. Now, the Figure 3 is reduced as shown in Figure 5. Apply voltage division rule in Figure 5 to find V Th . V Th = ( 40 s ) ( 10 s 10 s + 1000 ) = ( 40 s ) ( 10 s 10 ( s + 100 ) ) = 40 s + 100 Now, the Thevenin equivalent circuit is drawn as shown in Figure 6. Refer to Figure 6, the Thevenin impedance and the impedance of capacitor is connected in series. For the series connection, the voltage is same. The current ( I ) is calculated as follows: I = V Th Z Th + 5 × 10 5 s Substitute 40 s + 100 for V Th , and 2000 ( s + 50 ) s + 100 for Z Th in above equation to find I . I = ( 40 s + 100 ) 2000 ( s + 50 ) s + 100 + 5 × 10 5 s = ( 40 s + 100 ) 2000 s ( s + 50 ) + 5 × 10 5 ( s + 100 ) s ( s + 100 ) = 40 s 2000 s ( s + 50 ) + 5 × 10 5 ( s + 100 ) = 40 s 2000 s 2 + 1 × 10 5 s + 5 × 10 5 s + 5 × 10 7 Simplify the above equation as follows: I = 40 s 2000 s 2 + 6 × 10 5 s + 5 × 10 7 = 40 s 2000 ( s 2 + 300 s + 25000 ) I = 0.02 s s 2 + 300 s + 25000 (4) Write an expression to solve quadratic equation. s 1 , 2 = − b ± b 2 − 4 a c 2 (5) Here, a is the coefficient of second order term, b is the coefficient of first order term, and c is the coefficient of constant term. From equation (4), the characteristic equation of denominator is written as follows: s 2 + 300 s + 25000 = 0 Compare the above equation with the quadratic equation ( a s 2 + b s + c = 0 ) . a = 1 b = 300 c = 25000 Substitute 1 for a , 300 for b , and 25000 for c in equation (5) to find s 1 , 2 . s 1 , 2 = − 300 ± ( 300 ) 2 − 4 ( 1 ) ( 25000 ) 2 ( 1 ) = − 300 ± 90000 − 100000 2 = − 300 ± − 10000 2 = − 300 ± j 100 2 Simplify the equation. s 1 , 2 = − 300 + j 100 2 − 300 − j 100 2 = − 150 + j 50 , − 150 − j 50 Therefore, the roots of characteristic equations are real and imaginary. s 1 = − 150 + j 50 s 2 = − 150 − j 50 Now, the equation (4) is written as follows: I = 0.02 s ( s − ( − 150 + j 50 ) ) ( s − ( − 150 + j 50 ) ) = 0