Chapter 6 Practice Problems — Transient Analysis of RC and RL Circuits
These problems are meant to build your problem-solving intuition. A hint is provided for each — try the problem on your own first before reading it.
Problem 1 — RC Charging Time Constant
A 12 V DC source is connected to a series RC circuit with R = 22 kΩ and C = 47 μF at t = 0. The capacitor is initially uncharged.
(a) Calculate the time constant τ.
(b) Write the expression for \(V_C(t)\) for \(t \geq 0\).
(c) At what time has the capacitor charged to 10 V?
Hint
(a) The time constant of an RC circuit is simply \(\tau = RC\). Multiply your values carefully, keeping track of units (kΩ × μF = ms).
(b) For a capacitor charging from 0 to a final value \(V_\infty\), the universal step-response formula gives: [V_C(t) = V_\infty\left(1 - e^{-t/\tau}\right)]
(c) Set \(V_C(t) = 10\) V, substitute your expression, then solve for t by taking the natural logarithm of both sides. Remember \(\ln(e^x) = x\).
Problem 2 — RL Circuit Natural Response
An RL circuit has L = 50 mH and R = 200 Ω. The inductor carries an initial current of 80 mA when the source is disconnected at t = 0 (replaced by a short circuit through R).
(a) Find the time constant τ.
(b) Write the expression for \(i_L(t)\) for \(t \geq 0\).
(c) How long until the current falls below 5 mA?
Hint
(a) For an RL circuit, \(\tau = L/R\). Check units: mH/Ω = ms.
(b) With no driving source (natural response only), the inductor current decays exponentially from its initial value: [i_L(t) = i_L(0)\,e^{-t/\tau}]
(c) Set \(i_L(t) = 5\) mA and solve. You will need to take \(\ln\) of both sides. Think about the ratio \(5/80\) before computing.
Problem 3 — Complete Response with Non-Zero Initial Condition
In an RC circuit, R = 10 kΩ and C = 1 μF. At t = 0, a 20 V step is applied. The capacitor voltage just before switching is \(V_C(0^-) = 8\) V.
(a) Identify \(x(0)\), \(x(\infty)\), and τ.
(b) Use the universal formula to write \(V_C(t)\).
(c) At t = 2τ, what is \(V_C\)? What percentage of the total change has occurred?
Hint
(a) For the universal step-response formula \(x(t) = x(\infty) + [x(0) - x(\infty)]e^{-t/\tau}\): - \(x(0)\) is the value just after switching (capacitor voltage cannot jump, so \(V_C(0^+) = V_C(0^-)\)) - \(x(\infty)\) is the DC steady-state value long after switching (capacitor is an open circuit at DC)
(b) Substitute your three values directly into the formula.
(c) At \(t = 2\tau\), compute \(e^{-2}\). The percentage of total change is \((1 - e^{-2}) \times 100\%\).
Problem 4 — Finding Initial and Final Conditions
The circuit below is at DC steady state before t = 0. At t = 0, a switch opens, disconnecting a 5 kΩ parallel resistor.
Given: \(V_s = 15\) V, \(R_1 = 3\) k Ω, \(R_2 = 5\) kΩ (switched out at t=0), \(C = 10\) μF.
(a) Find \(V_C(0^-)\) (before switching).
(b) Find \(V_C(\infty)\) (long after switching).
(c) Find the new τ after switching.
Hint
(a) In DC steady state, the capacitor is an open circuit. Use a voltage divider with \(R_1\) and \(R_2\) (both present) to find \(V_C(0^-)\).
(b) After the switch opens, \(R_2\) is removed. Now find the new DC steady state — again treat the capacitor as open and redo the voltage divider with only \(R_1\) remaining.
(c) The Thévenin resistance seen by the capacitor after switching is just \(R_1\) (since \(R_2\) is disconnected). Use \(\tau = R_{th} \cdot C\).
Problem 5 — Energy Stored and Dissipated
An RC circuit has R = 1 kΩ and C = 100 μF. The capacitor is initially charged to 20 V and then discharges through R.
(a) How much energy was initially stored in the capacitor?
(b) Write the expression for \(V_C(t)\) during the discharge.
(c) How much energy has been dissipated in R by t = τ?
(d) What fraction of the initial energy remains in C at t = τ?
Hint
(a) Energy stored in a capacitor: \(W = \frac{1}{2}CV^2\). Substitute \(V = 20\) V.
(b) Discharging from \(V_0\) with no source is a natural response: \(V_C(t) = V_0\,e^{-t/\tau}\).
(c) Energy remaining in C at time t is \(\frac{1}{2}C[V_C(t)]^2\). Energy dissipated = initial energy − remaining energy. Compute \(V_C(\tau)\) first.
(d) Divide the energy remaining at t = τ by the initial energy. The answer is a clean expression in terms of \(e\).