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Thevenin: current through 10 Ω

Short answer

Thevenin's theorem: Vth = 10V + (5A×15Ω) = 85V, Rth = 15Ω (current source opened). Current through 10Ω is I = 85/(15+10) = 3.4 A.

Determine the current through the 10 Ω resistance using Thevenin's theorem. [Figure: 10 V source in series with 15 Ω, then 10 Ω resistor in a vertical branch, then 5 Ω in series, and a 5 A current source in the last branch.]

Assumed circuit: the 10 V source with 15 Ω in series connects to node A. The 10 Ω load is the vertical branch from A to ground. The last branch is the 5 A current source with a 5 Ω resistor in series, pushing 5 A into node A (direction taken as upward toward A).

Step 1: remove the 10 Ω and find Vth

With the load open, the 5 A from the current source must return through the 15 Ω and the 10 V source. The 5 Ω is in series with an ideal current source, so it has no effect on the voltage at A.

Vth = 10 + 5 × 15 = 85 V

Step 2: find Rth

Short the 10 V source and open the current source (its branch, including the 5 Ω, becomes open):

Rth = 15 Ω

Step 3: load current

I = Vth/(Rth + RL) = 85/(15 + 10) = 85/25

Answer: I = 3.4 A through the 10 Ω resistor. (If the current source points the other way, Vth = 10 − 75 = −65 V and I = −2.6 A.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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