(a) Explanation Given Data: New value of resistor is 1 k Ω and The function for current is given as follows: i 1 = 5 u ( t ) − 4 mA . Formula used: The expression for the exponential damping coefficient or the neper frequency for series R L C circuit is as follows: α = R 2 L (1) Here, α is the exponential damping coefficient or the neper frequency of series R L C circuit, R is the resistance of a series R L C circuit, L is the inductance of a series R L C circuit. The expression for the resonant frequency is as follows: ω 0 = 1 L C (2) Here, ω 0 is the resonant frequency and L is the inductance of a parallel R L C circuit. The expression for the two solutions of the characteristic equation of a series R L C circuit is as follows: s 1 = − α + α 2 − ω 0 2 (3) s 2 = − α − α 2 − ω 0 2 (4) Here, s 1 and s 2 is the solutions of the characteristic equation of a series R L C circuit. The expression for complete natural response for source free series R L C circuit is: i L = i f + i n (5) Here, i L is current through inductor for t > 0 , i f is forced response and i n is the natural response. The expression for natural response for source free series R L C circuit is: i n = A 1 e S 1 t + A 2 e S 2 t (6) Here, A 1 and A 2 are arbitrary constants and t is the time. Calculation: The unit-step forcing function as a function of time which is zero for all values of its argument less than zero and which is unity for all positive values of its argument. u ( t − t 0 ) = { 0 t < t 0 1 t > t 0 Here, t 0 is the time at which the argument of the function becomes zero. Substitute 0 − for t in given function. i 1 = 5 u ( 0 − ) − 4 mA = 5 ( 0 ) − 4 mA = − 4 mA The capacitor and the inductor are connected in the circuit for long time. So, the capacitor behaves as open circuit and the inductor behaves as short circuit. The redrawn circuit diagram is given in Figure 1 for t = 0 − . Refer to the redrawn Figure 1: Series connected branch have same value of current . and as short circuited inductor is in series with current source i 1 . So, i L ( 0 − ) = − 4 mA Inductor does not allow sudden change in current. So, i L ( 0 + ) = i L ( 0 − ) = − 4 mA The expression for v R ( 0 − ) is as follows: v R ( 0 − ) = i 1 R (7) Here, v R ( 0 − ) is voltage across resistor for t = 0 − , i 1 is current source and R is the resistance in the circuit. Substitute − 4 mA for i 1 and 1 k Ω for R in equation (7). v R ( 0 − ) = ( − 4 mA ) ( 1 k Ω ) = ( − 4 × 10 − 3 A ) ( 1 × 10 3 Ω ) { ∵ 1 mA = 10 − 3 A ∵ 1 k Ω = 10 3 Ω } = − 4 V As parallel branch have same voltage and open circuited terminal of capacitor is in parallel to resistor. So, v C ( 0 − ) = − 4 V The capacitor does not allow sudden change in the voltage. So, v C ( 0 + ) = v C ( 0 − ) = − 4 V Substitute ∞ for t in given function. i 1 = 5 u ( ∞ ) − 4 mA = 5 ( 1 ) − 4 mA = 1 mA The capacitor and the inductor are connected in the circuit for long time. So, the capacitor behaves as open circuit and the inductor behaves as short circuit. The redrawn circuit diagram is given in Figure 2 for t = 0 − . Refer to the redrawn Figure 2: Series connected branch have same value of current . and as short circuited inductor is in series with current source i S . So i L ( ∞ ) = 1 mA Refer to the FIGURE 9.55 given in the textbook. Substitute 1 k Ω for R and 2 mH for L in equation (1). α = 1 k Ω 2 ( 2 mH ) = 1 × 10 3 Ω 2 ( 2 × 10 − 3 H ) { ∵ 1 k Ω = 10 3 Ω ∵ 1 mH = 10 − 3 H } = 2.5 × 10 5 s − 1 Substitute 2 mH for L and 20 nF for C in equation (2). ω 0 = 1 ( 2 mH ) ( 20 nF ) = 1 ( 2 × 10 − 3 H ) ( 20 × 10 − 9 F ) { ∵ 1 mH = 10 − 3 H ∵ 1 nF = 10 − 9 F } = 1 6.324 × 10 − 6 rad s = 1.58 × 10 5 rad s As value of neper frequency α is greater than resonant frequency ω 0 , therefore, circuit is over-damped. Substitute 2.5 × 10 5 s − 1 for α and 1.58 × 10 5 rad s for ω 0 in equation (3). s 1 = − 2.5 × 10 5 s − 1 + ( 2.5 × 10 5 s − 1 ) 2 − ( 1.58 × 10 5 rad s ) 2 = − 2.5 × 10 5 s − 1 + 6.25 × 10 10 s − 2 − 2.49 × 10 10 ( r a d s ) 2 = − 2.5 × 10 5 s − 1 + 1.94 × 10 5 s − 1 = − 0.56 × 10 5 s − 1 Substitute 2.5 × 10 5 s − 1 for α and 1.58 × 10 5 rad s for ω 0 in equation (4). s 2 = − 2.5 × 10 5 s − 1 − ( 2.5 × 10 5 s − 1 ) 2 − ( 1.58 × 10 5 rad s ) 2 = − 2.5 × 10 5 s − 1 − 6.25 × 10 10 s − 2 − 2.49 × 10 10 ( r a d s ) 2 = − 2.5 × 10 5 s − 1 − 1.94 × 10 5 s − 1 = − 4.44 × 10 5 s − 1 Forced response is the value of current for t = ∞ (b) To determine Plot the graph for 0 ≤ t ≤ 200 μ s