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Norton, R for maximum power, 48 V and 5 A

Short answer

Norton equivalent: I_N = 9 A, R_N = 4 Ω. For maximum power, R = R_N = 4 Ω. Maximum power P_max = I_N²·R_N/4 = 81 W.

Use Norton's theorem to calculate the value of R that will absorb maximum power from the circuit shown in the figure below. Also calculate the maximum power drawn by it. [Figure: 48 V source in series with 12 Ω on top; R in a vertical branch; 6 Ω shunt; 5 A current source.]

Reading the figure

From left to right: 48 V battery (+ on top) in series with 12 Ω; the variable resistor R; a 6 Ω shunt; a 5 A source (arrow up) feeding the top rail. R is the load.

Step 1: Norton current (short R)

With R replaced by a short circuit, the 6 Ω shunt is also shorted.

  • Current from the 48 V branch: 48/12 = 4 A (towards the short)
  • The 5 A source also feeds the short.

I_N = 4 + 5 = 9 A

Step 2: Norton resistance (sources off)

48 V shorted and 5 A opened: R_N = 12 ∥ 6 = 12 × 6/18 = 4 Ω

Step 3: Value of R for maximum power

R = R_N = 4 Ω

Step 4: Maximum power

Current in R = I_N/2 = 4.5 A P_max = I² R = 4.5² × 4 = 81 W (also P_max = I_N² R_N/4 = 81 × 4/4 = 81 W)

Check with Thevenin: V_th = I_N R_N = 36 V, P_max = V_th²/(4 R_N) = 1296/16 = 81 W.

Answer: R = 4 Ω; maximum power = 81 W

✓ Answer checked by Claude Opus 5.5 (high effort)

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