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Thevenin theorem, power in 10Ω

Short answer

Open-circuit voltage across 10Ω terminals is 8.29 V, Thevenin resistance is 4.327 Ω. Current through 10Ω = 8.29/(4.327+10) = 0.5787 A. Power dissipated = I²×10 = 3.35 W.

Find the power dissipated in 10 Ohm resistor using Thevenin theorem of the following network. Note that all resistances are in Ohm. [Figure: 5, 6, 8 and 2 Ω resistors around the outside with a 10 Ω resistor in the middle vertical branch; 24 V source at the bottom.]

Reading the figure

The 24 V source is connected between the left node P and the right node Q. Two paths join P to Q: P–5 Ω–T–6 Ω–Q (upper) and P–8 Ω–M–2 Ω–Q (lower). The 10 Ω load joins T and M.

   P --[5]-- T --[6]-- Q
   |         |         |
   |       [10]        |
   |         |         |
   P --[8]-- M --[2]-- Q      (24 V source between P and Q)

The polarity of the source does not change the power in the 10 Ω. Take V_P = 0 and V_Q = 24 V.

Step 1: Thevenin voltage (10 Ω removed)

V_T = 24 × 5/(5 + 6) = 10.909 V V_M = 24 × 8/(8 + 2) = 19.2 V

V_th = V_M − V_T = 8.29 V

Step 2: Thevenin resistance (source shorted, so P = Q)

From T: 5 Ω ∥ 6 Ω = 2.727 Ω From M: 8 Ω ∥ 2 Ω = 1.6 Ω

R_th = 2.727 + 1.6 = 4.327 Ω

Step 3: Current and power in the 10 Ω

I = V_th/(R_th + 10) = 8.291/14.327 = 0.5787 A

P = I² × 10 = 0.5787² × 10 = 3.35 W

Answer: Power dissipated in the 10 Ω = 3.35 W

✓ Answer checked by Claude Opus 5.5 (high effort)

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