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Chapter 6 Quiz — Transient Analysis of RC and RL Circuits

Multiple Choice Questions

1. An RC circuit has R = 47 kΩ and C = 10 μF. What is the time constant?

  • A) 47 ms
  • B) 470 ms
  • C) 4.7 s
  • D) 4.7 ms
Answer

B) 470 ms

\(\tau = RC = 47 \times 10^3 \times 10 \times 10^{-6} = 470 \times 10^{-3}\) s = 470 ms.

2. A capacitor is initially uncharged and is connected to a 10 V source through a resistor. After one time constant, the capacitor voltage is approximately:

  • A) 3.68 V
  • B) 6.32 V
  • C) 5 V
  • D) 10 V
Answer

B) 6.32 V

After \(t = \tau\): \(V_C = V_S(1 - e^{-1}) = 10(1 - 0.368) = 10 \times 0.632 = 6.32\) V.

3. How many time constants must elapse before a first-order circuit is considered to have reached steady state (99.3% complete)?

  • A) 1
  • B) 3
  • C) 5
  • D) 10
Answer

C) 5

After \(5\tau\): \(e^{-5} \approx 0.0067\), so the circuit is 99.3% of the way to its final value. Engineers consider the transient complete after 5 time constants.

4. An inductor cannot change its current instantaneously. This is because:

  • A) Resistors in series limit the current
  • B) An instantaneous current change would require infinite voltage (\(v = L\,di/dt\))
  • C) Capacitors block the change
  • D) KVL prevents it
Answer

B) An instantaneous current change would require infinite voltage

The V-I relationship \(v = L\,di/dt\) means that \(di/dt \to \infty\) (instantaneous change) would require \(v \to \infty\), which is physically impossible in a real circuit.

5. For an RL circuit with L = 100 mH and R = 50 Ω, what is the time constant?

  • A) 5 s
  • B) 50 ms
  • C) 2 ms
  • D) 0.5 s
Answer

C) 2 ms

\(\tau = L/R = 0.1 / 50 = 0.002\) s = 2 ms.

6. Using the universal step-response formula, what value does \(x(t)\) approach as \(t \to \infty\)?

  • A) \(x(0)\)
  • B) Zero
  • C) \(x(\infty)\)
  • D) \(\tau\)
Answer

C) \(x(\infty)\)

The formula is \(x(t) = x(\infty) + [x(0) - x(\infty)]e^{-t/\tau}\). As \(t \to \infty\), \(e^{-t/\tau} \to 0\), so \(x(t) \to x(\infty)\).

7. When analyzing a complex circuit to find the time constant \(\tau\), what do you use as the resistance \(R\)?

  • A) The total resistance in the circuit
  • B) The Thévenin resistance seen by the energy storage element (C or L)
  • C) The series resistance only
  • D) The parallel resistance only
Answer

B) The Thévenin resistance seen by the energy storage element

Kill all independent sources, remove the capacitor or inductor, and find the resistance seen at those terminals — this is \(R_{Th}\). Then \(\tau = R_{Th} C\) or \(\tau = L/R_{Th}\).

8. In DC steady state, how does an inductor behave?

  • A) As an open circuit with large voltage across it
  • B) As a short circuit with zero voltage across it
  • C) As a resistor
  • D) As a capacitor
Answer

B) As a short circuit with zero voltage across it

In DC steady state, current is constant (\(di/dt = 0\)), so \(v_L = L\,di/dt = 0\). An inductor is a short circuit (a wire) at DC.

9. A capacitor voltage is \(V_C(t) = 15 - 15e^{-2t}\) V. What are the initial and final values?

  • A) Initial = 0 V, Final = 15 V
  • B) Initial = 15 V, Final = 0 V
  • C) Initial = 0 V, Final = 30 V
  • D) Initial = 7.5 V, Final = 15 V
Answer

A) Initial = 0 V, Final = 15 V

At \(t = 0\): \(V_C(0) = 15 - 15e^0 = 15 - 15 = 0\) V.
As \(t \to \infty\): \(V_C(\infty) = 15 - 0 = 15\) V.

10. The "natural response" of a first-order circuit is:

  • A) The steady-state response to the source
  • B) The circuit's response due to initial stored energy, which decays to zero
  • C) The complete response including all sources
  • D) The sinusoidal AC response
Answer

B) The circuit's response due to initial stored energy, which decays to zero

The natural response \(x_n(t) = Ae^{-t/\tau}\) results from the energy initially stored in the capacitor or inductor, decaying exponentially to zero without any external forcing.

Practice Problems

Problem 1: RC Charging Transient

A series RC circuit has R = 20 kΩ, C = 25 μF, and a 15 V source. The switch closes at \(t = 0\) with the capacitor initially uncharged.

a) Find the time constant.
b) Write the expression for \(V_C(t)\).
c) What is \(V_C\) after 1 time constant? After 3 time constants?
d) How long until \(V_C = 10\) V?

Solution

a) \(\tau = RC = 20\,000 \times 25 \times 10^{-6} = 0.5\) s

b) \(V_C(t) = 15(1 - e^{-t/0.5}) = 15(1 - e^{-2t})\) V

c) After \(\tau = 0.5\) s: \(V_C = 15 \times 0.632 = 9.48\) V

After \(3\tau = 1.5\) s: \(V_C = 15(1 - e^{-3}) = 15 \times 0.950 = 14.25\) V

d) Solve \(10 = 15(1 - e^{-2t})\):

\(e^{-2t} = 1 - 10/15 = 1/3\)

\(-2t = \ln(1/3) = -\ln 3\)

\(t = \frac{\ln 3}{2} = \frac{1.099}{2} \approx 0.55\) s

Problem 2: RL Energizing Transient

A 12 V source is switched into an RL circuit with R = 6 Ω and L = 30 mH at \(t = 0\). The inductor carries no initial current.

a) Find the time constant and the final (steady-state) current.
b) Write the expression for \(i_L(t)\).
c) Find the inductor voltage \(v_L(t)\).
d) Verify that at \(t = 0^+\): the inductor voltage equals 12 V and current is 0.

Solution

a) \(\tau = L/R = 30 \times 10^{-3} / 6 = 5\) ms

Final current: \(I_\infty = V_S/R = 12/6 = 2\) A

b) \(i_L(t) = 2(1 - e^{-t/0.005}) = 2(1 - e^{-200t})\) A

c) \(v_L = L\,di/dt = 0.03 \times 2 \times 200\,e^{-200t} = 12\,e^{-200t}\) V

d) At \(t = 0^+\): - \(i_L(0) = 2(1-1) = 0\) A ✓ (current starts at zero) - \(v_L(0) = 12\,e^0 = 12\) V ✓ (all source voltage appears across inductor initially)

Problem 3: Universal Step-Response Formula

A circuit contains a 20 V source and two resistors: R1 = 4 Ω, R2 = 6 Ω. A capacitor C = 500 μF is in parallel with R2. The switch closes at \(t = 0\). Before switching, the capacitor was charged to 5 V.

a) Find \(V_C(0)\), \(V_C(\infty)\), and \(\tau\).
b) Write \(V_C(t)\).
c) At what time does \(V_C = 10\) V?

Solution

a) \(V_C(0) = 5\) V (given)

\(V_C(\infty)\): At DC steady state, C is open. Voltage divider: \(V_C(\infty) = 20 \times \frac{6}{4+6} = 12\) V

\(\tau\): Kill source (short it). R1 and R2 in parallel as seen by C: \(R_{Th} = \frac{4 \times 6}{4+6} = 2.4\) Ω
\(\tau = R_{Th}C = 2.4 \times 500 \times 10^{-6} = 1.2\) ms

b) \(V_C(t) = 12 + (5 - 12)e^{-t/0.0012} = 12 - 7e^{-833t}\) V

c) Solve \(10 = 12 - 7e^{-833t}\):

\(7e^{-833t} = 2\)
\(e^{-833t} = 2/7\)
\(t = -\ln(2/7)/833 = \ln(7/2)/833 = 1.253/833 \approx 1.5\) ms

Problem 4: RC Discharge with Energy

A 100 μF capacitor is charged to 24 V and then discharged through a 2 kΩ resistor at \(t = 0\).

a) Find the initial stored energy.
b) Write \(V_C(t)\) and \(i(t)\).
c) Find the power dissipated in the resistor as a function of time.
d) Show that the total energy dissipated equals the initial stored energy.

Solution

a) \(E_0 = \frac{1}{2}CV^2 = \frac{1}{2} \times 100\times10^{-6} \times 576 = 28.8\) mJ

b) \(\tau = RC = 2000 \times 100 \times 10^{-6} = 0.2\) s

\(V_C(t) = 24\,e^{-5t}\) V (since \(1/\tau = 5\))

\(i(t) = \frac{V_C}{R} = \frac{24}{2000}e^{-5t} = 12\,e^{-5t}\) mA

c) \(p_R(t) = i^2 R = (12\times10^{-3})^2 \times 2000 \times e^{-10t} = 0.288\,e^{-10t}\) W

d) Total energy dissipated:

\(E = \int_0^\infty 0.288\,e^{-10t}\,dt = 0.288 \times \frac{1}{10} = 0.0288\) J = 28.8 mJ ✓

All stored energy is dissipated as heat in the resistor.