- 2073 Shrawan · 8 marks
Find R for maximum power to load
Short answer
Thevenin resistance seen by RL is 2 + (R∥4). Set it equal to RL = 4 Ω, solving gives R = 4 Ω. Thevenin voltage is 12.67 V, so maximum power in RL is V_th²/(4·4) = 10.03 W.
Find the value of Resistance R such that the load resistance RL which is equal to 4 Ω, will deliver maximum power. Also find that maximum power. [Figure: 3 A current source on the left; 6 Ω shunt; 20 V source in the top branch; R in series with a 4 Ω resistor in parallel; 3 Ω shunt; load RL = 4 Ω.]
Reading the figure
Bottom rail = ground. Left node P: 3 A source (up) into P, 6 Ω to ground, and the 20 V battery (+ on the right) to node N. Node N: 3 Ω to ground, and two parallel resistors R and 4 Ω from N to the top of the load R_L = 4 Ω.
Step 1: Thevenin resistance seen by R_L
Sources off (3 A opened, 20 V shorted): P is joined to N, so the 6 Ω and 3 Ω are in parallel at N: 6 ∥ 3 = 2 Ω
Looking into R_L: R_th = 2 + (R ∥ 4)
Step 2: Matching condition
For maximum power in R_L, R_th = R_L = 4 Ω:
2 + R × 4/(R + 4) = 4 → 4R/(R + 4) = 2 → 4R = 2R + 8 → R = 4 Ω
Step 3: Thevenin voltage (R_L removed)
No current flows in R or the 4 Ω, so V_th = V_N. V_N = V_P + 20, and 3 = V_P/6 + V_N/3
3 = V_P/6 + (V_P + 20)/3 → V_P/2 = 3 − 6.667 → V_P = −7.33 V V_th = V_N = 12.67 V
Step 4: Maximum power
P_max = V_th²/(4 R_L) = 12.67²/(4 × 4) = 160.4/16 = 10.03 W
Answer: R = 4 Ω; P_max ≈ 10.03 W
Note: this is the usual matching treatment (R_th seen by the load equals R_L). Strictly, with R_L fixed, the load power would be even larger for a smaller R, so the question is read as a matching problem.
✓ Answer checked by Claude Opus 5.5 (high effort)
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