Explanation The given circuit is shown in Figure 1. To find out the load impedance R L for maximum power transfer condition, Thevenin equivalent referred to the load impedance is required. Determine the open circuit voltage E O C . The required diagram is shown in Figure 2. The conversion from k Ω to Ω is given by, 1 k Ω = 1000 Ω Hence, the resistance R 1 is, R 1 = 1000 Ω Apply KVL in the circuit as, − 1 ∠ 0 ° + I × 1000 Ω = 0 I = 1 ∠ 0 ° 1000 I = 0.001 ∠ 0 ° A The conversion from A to mA is given by, 1 A = 10 3 mA The conversion from 0 .001 A to mA is given by, 0 .001 A = 0.001 × 10 3 mA = 1 mA Hence, the current I is, I = 1 ∠ 0 ° mA The conversion from 40 k Ω to Ω is given by, 40 k Ω = 40000 Ω Hence, the resistance R 2 is, R 2 = 40000 Ω The open circuit voltage E O C from the circuit is calculated as, E O C = − 50 I R 2 Substitute 40000 Ω for R 2 and 0.001 ∠ 0 ° A for I in the above equation. E O C = − 50 ( 0.001 ∠ 0 ° A ) ( 40000 Ω ) = − 2000 ∠ 0 ° V Short circuit the load branch and determine the short circuit current flowing through it. The required diagram is shown in Figure 3. Since the resistance R 2 is short circuited hence it can be neglected. The required diagram is shown in Figure 4. The current I remains as it is. The short circuit current I S C is given by, I S C = − 50 I Substitute 0

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN: 9780133923605

Author: Robert L. Boylestad

Publisher: PEARSON

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Chapter 19, Problem 52P

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The load impedance RL and the maximum power delivered to RL for the given condition.

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Chapter 19, Problem 51P

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