Chapter 4 Practice Problems

Practice Problems

Problem 1: Thévenin Equivalent

A circuit has a 30 V source, with R1 = 10 Ω in series with the source, and R2 = 15 Ω connected between node A and ground. Find the Thévenin equivalent circuit seen at terminals A–B (where B is ground).

Solution

Finding \(V_{Th}\):

Remove any load from terminals A–B (open circuit). Voltage at A:

R1 and R2 form a voltage divider: \(V_{Th} = 30 \times \frac{15}{10+15} = 30 \times 0.6 = 18\) V

Finding \(R_{Th}\):

Kill the 30 V source (replace with short). Looking into terminals A–B: R1 and R2 are in parallel: \(R_{Th} = \frac{10 \times 15}{10+15} = \frac{150}{25} = 6\) Ω

Thévenin equivalent: 18 V source in series with 6 Ω.

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Problem 2: Maximum Power Transfer

Using the Thévenin equivalent found in Problem 1, find:

a) The load resistance that maximizes power transfer
b) The maximum power delivered to the load
c) The efficiency at maximum power transfer

Solution

a) \(R_L = R_{Th} = 6\) Ω

b) \(P_{max} = \frac{V_{Th}^2}{4 R_{Th}} = \frac{18^2}{4 \times 6} = \frac{324}{24} = 13.5\) W

c) Efficiency = 50% (always at maximum power transfer condition).

Verification: With \(R_L = 6\) Ω, current \(I = 18/(6+6) = 1.5\) A.
Power to load: \(P_L = 1.5^2 \times 6 = 13.5\) W ✓
Power from source: \(P_S = 18 \times 1.5 = 27\) W
Efficiency: \(13.5/27 = 50\%\) ✓

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Problem 3: Source Transformation

Simplify the following circuit using source transformation: a 9 V voltage source with a 3 Ω series resistor in parallel with a 6 Ω resistor, driving a 4 Ω load.

Solution

Step 1: Transform the 9 V/3 Ω voltage source to a current source: \(I_N = 9/3 = 3\) A in parallel with 3 Ω.

Step 2: The 3 Ω and 6 Ω are now in parallel: \(R_{eq} = (3 \times 6)/(3+6) = 2\) Ω

Step 3: The circuit is now 3 A current source in parallel with 2 Ω and 4 Ω.

Load current (current divider): \(I_L = 3 \times \frac{2}{2+4} = 1\) A
Load voltage: \(V_L = 1 \times 4 = 4\) V

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Problem 4: Norton Equivalent with Verification

Find the Norton equivalent at terminals A–B for a circuit with a 24 V source, R1 = 8 Ω in series with the source, and R2 = 4 Ω from node A to ground. Verify using \(R_N = V_{oc}/I_{sc}\).

Solution

\(I_{sc}\) (short A to B):

R2 is shorted. The only resistance is R1: \(I_{sc} = 24/8 = 3\) A
This is \(I_N = 3\) A.

\(R_N\) (kill source):

Short the 24 V source. Looking into A–B: R1 ∥ R2 = (8×4)/(8+4) = 32/12 = 8/3 Ω ≈ 2.67 Ω.

Verification via \(V_{oc}/I_{sc}\):

\(V_{oc}\) = voltage at A with B open = \(24 \times 4/(8+4) = 8\) V
\(R_{Th} = V_{oc}/I_{sc} = 8/3\) Ω ✓

Norton equivalent: 3 A current source in parallel with 8/3 Ω (≈ 2.67 Ω).