Chapter 4 Practice Problems — DC Circuit Analysis Methods
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 — Source Transformation
A practical voltage source is modeled as a 24 V ideal voltage source in series with an internal resistance of \(R_s = 6\) Ω.
(a) Transform this voltage source model into an equivalent Norton current source in parallel with a resistor. State the current source value and the parallel resistance.
(b) Connect a load \(R_L = 12\) Ω to the Norton equivalent and find the load current and load voltage.
(c) Verify your answer by returning to the original Thévenin (series) model and computing the same quantities using a simple voltage divider.
Hint
(a) Norton current: \(I_N = V_S / R_S = 24 / 6\). The Norton resistance equals the series source resistance: \(R_N = R_S = 6\) Ω.
(b) In the Norton equivalent, \(R_N\) and \(R_L\) are in parallel. Use the current divider: \(I_L = I_N \cdot \frac{R_N}{R_N + R_L}\). Then \(V_L = I_L R_L\).
(c) In the Thévenin model, total resistance is \(R_S + R_L\). Use \(I = 24 / (R_S + R_L)\) and \(V_L = I R_L\). Both methods must give identical results.
Problem 2 — Thévenin Equivalent Circuit
Find the Thévenin equivalent seen by a load connected between terminals A and B, given: a 30 V source, \(R_1 = 10\) Ω in series with the source, and \(R_2 = 20\) Ω connected from the junction of \(R_1\) to ground (node B).
(a) Find the open-circuit voltage \(V_{Th}\) by computing the voltage at node A with no load connected.
(b) Find \(R_{Th}\) by killing the independent source (replacing the 30 V source with a short circuit) and computing the resistance looking into terminals A–B.
(c) A load \(R_L = 10\) Ω is connected from A to B. Using the Thévenin equivalent, find the load voltage and current.
Hint
(a) With no load, \(R_1\) and \(R_2\) form a voltage divider. \(V_{Th} = V_{oc} = 30 \times \frac{R_2}{R_1 + R_2}\).
(b) Short the source. Looking into A–B, you see \(R_2\) in parallel with \(R_1\) (the shorted source connects the far end of \(R_1\) to ground). \(R_{Th} = R_1 \| R_2\).
(c) In the Thévenin equivalent, \(V_{Th}\) drives \(R_{Th}\) in series with \(R_L\). Apply Ohm's Law: \(I_L = V_{Th} / (R_{Th} + R_L)\) and \(V_L = I_L R_L\).
Problem 3 — Norton Equivalent and Maximum Power Transfer
A network has a Thévenin equivalent of \(V_{Th} = 20\) V and \(R_{Th} = 25\) Ω.
(a) Determine the Norton equivalent current \(I_N\) and resistance \(R_N\).
(b) What value of load resistance \(R_L\) achieves maximum power transfer from the source to the load?
(c) Calculate the maximum power delivered to \(R_L\) and the efficiency of power transfer at this condition.
Hint
(a) Norton current: \(I_N = V_{Th} / R_{Th}\). Norton resistance equals Thévenin resistance: \(R_N = R_{Th}\).
(b) The maximum power transfer theorem states that maximum power is delivered when \(R_L = R_{Th}\).
(c) Maximum power: \(P_{max} = V_{Th}^2 / (4 R_{Th})\). At maximum transfer, the efficiency is exactly 50% — half the source power goes to the load, half is dissipated internally in \(R_{Th}\).
Problem 4 — Nodal Analysis
A circuit has three nodes: reference (ground), node 1, and node 2. Components: - 10 V source from node 1 to ground - \(R_1 = 5\) Ω from node 1 to node 2 - \(R_2 = 10\) Ω from node 2 to ground - \(R_3 = 20\) Ω from node 2 to ground
(a) Identify the node voltages. Note that the voltage source fixes one node voltage directly.
(b) Write the KCL equation at node 2, expressing all currents leaving the node in terms of node voltages.
(c) Solve for \(V_2\) and find the current through each resistor.
Hint
(a) Since a 10 V source connects node 1 to ground, \(V_1 = 10\) V by inspection. Only \(V_2\) is unknown.
(b) KCL at node 2 (currents leaving): \(\frac{V_2 - V_1}{R_1} + \frac{V_2}{R_2} + \frac{V_2}{R_3} = 0\). Substitute \(V_1 = 10\) V.
(c) Collect terms and solve for \(V_2\). Then apply Ohm's Law to each resistor using the node voltage difference across it.