For the circuit of Figure 3, the parameters are: I; = 2A, L = 5mH, C = 5 mF, R, = 102, R1 = 1.81 2, and R2 = 0.2 2. After the switch has been in its initial position for a long time, at t = 0 it moves instantaneously to the left as shown in Figure 3. What is the initial value of the time derivative of the capacitor voltage in the circuit, dv(0")ldt ? |3D R1 t= 0 Rs R23 Figure 3 dv(0*) а) =-7500 V/s dt dv(0*) = 0 dt dv(0*) c) = 1,200 V/s dt dv(0*) d) =-2200 V/s dt e) None of the above For the circuit of Figure 3, identify the correct general expression for the voltage, v(t), for t2 0. a) v(t) = e5 (7.5 cos 45t +17 sin 60t) (V) b) v(t)=e¯55 (cos60t – 17 sin 60t ) (V) -255t ©) v(t) = 110.4e-1811 – 90.4e-2211 (V) d) v(t) = 75.2e-2211 e) None of the above - 55.2e-181 (V)

Introductory Circuit Analysis (13th Edition)
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For the circuit in Figure 3, the parameters are: \( I_s = 2A \), \( L = 5mH \), \( C = 5 mF \), \( R_s = 10\Omega \), \( R_1 = 1.81 \Omega \), and \( R_2 = 0.2 \Omega \). After the switch has been in its initial position for a long time, at \( t = 0 \) it moves instantaneously to the left as shown in Figure 3. What is the initial value of the time derivative of the capacitor voltage in the circuit, \( \frac{dv(0^+)}{dt} \)?

![Circuit Diagram](image)

**Options:**
a) \( \frac{dv(0^+)}{dt} = -7500 \, V/s \)

b) \( \frac{dv(0^+)}{dt} = 0 \)

c) \( \frac{dv(0^+)}{dt} = 1200 \, V/s \)

d) \( \frac{dv(0^+)}{dt} = -2200 \, V/s \)

e) None of the above

**For the circuit of Figure 3, identify the correct general expression for the voltage, \( v(t) \), for \( t \geq 0 \).**

a) \( v(t) = e^{-45t} (7.5\cos 45t + 17 \sin 60t) \, (V) \)

b) \( v(t) = e^{-25t} (\cos 60t - 17 \sin 60t) \, (V) \)

c) \( v(t) = 110.4e^{-18t} - 90.4e^{-22t} \, (V) \)

d) \( v(t) = 75.2e^{-22t} - 55.2e^{-18t} \, (V) \)

e) None of the above

**For the circuit in Figure 3, determine the final value of the inductor current, \( i_L(\infty) \).**

**Explanation:**

Figure 3 depicts a circuit with an inductor (\( L \)), a capacitor (\( C \)), two resistors (\( R_1 \), \(
Transcribed Image Text:For the circuit in Figure 3, the parameters are: \( I_s = 2A \), \( L = 5mH \), \( C = 5 mF \), \( R_s = 10\Omega \), \( R_1 = 1.81 \Omega \), and \( R_2 = 0.2 \Omega \). After the switch has been in its initial position for a long time, at \( t = 0 \) it moves instantaneously to the left as shown in Figure 3. What is the initial value of the time derivative of the capacitor voltage in the circuit, \( \frac{dv(0^+)}{dt} \)? ![Circuit Diagram](image) **Options:** a) \( \frac{dv(0^+)}{dt} = -7500 \, V/s \) b) \( \frac{dv(0^+)}{dt} = 0 \) c) \( \frac{dv(0^+)}{dt} = 1200 \, V/s \) d) \( \frac{dv(0^+)}{dt} = -2200 \, V/s \) e) None of the above **For the circuit of Figure 3, identify the correct general expression for the voltage, \( v(t) \), for \( t \geq 0 \).** a) \( v(t) = e^{-45t} (7.5\cos 45t + 17 \sin 60t) \, (V) \) b) \( v(t) = e^{-25t} (\cos 60t - 17 \sin 60t) \, (V) \) c) \( v(t) = 110.4e^{-18t} - 90.4e^{-22t} \, (V) \) d) \( v(t) = 75.2e^{-22t} - 55.2e^{-18t} \, (V) \) e) None of the above **For the circuit in Figure 3, determine the final value of the inductor current, \( i_L(\infty) \).** **Explanation:** Figure 3 depicts a circuit with an inductor (\( L \)), a capacitor (\( C \)), two resistors (\( R_1 \), \(
For the circuit in Figure 3 before the switching event, i.e., for \( t < 0 \), determine the steady-state value of the current flowing through the resistor, \( R \).
Transcribed Image Text:For the circuit in Figure 3 before the switching event, i.e., for \( t < 0 \), determine the steady-state value of the current flowing through the resistor, \( R \).
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