Skip to main content

Thevenin theorem, current in 10Ω

Short answer

Thevenin voltage is −6.286 V and Thevenin resistance is 9.429 Ω. The current in the 10 Ω resistor is 6.286/(9.429+10) ≈ 0.324 A, flowing upward.

For the circuit shown in figure below, calculate the current in the 10 ohm resistance using Thevenin's theorem. [Figure: 5 Ω resistor in series with a 20 V source on the left; 2 Ω shunt in the middle; 8 Ω in series with a 12 V source on the top right; 10 Ω resistor on the right.]

Reading the figure

Left branch: 5 Ω in series with the 20 V battery (+ on top) to the bottom rail. Middle: 2 Ω shunt from the top node N to the bottom rail. Right: 8 Ω in series with the 12 V battery (+ on the left, towards N), then down through the 10 Ω to the bottom rail. The 10 Ω is the load.

Step 1: Thevenin voltage (10 Ω removed)

No current in the 8 Ω or the 12 V battery, so

V_N = 20 × 2/(5 + 2) = 5.714 V (voltage divider of the 20 V source between 5 Ω and 2 Ω)

The 12 V battery has its + on the N side, so the open terminal is 12 V lower:

V_th = 5.714 − 12 = −6.286 V

The load terminal is 6.286 V below the bottom rail.

Step 2: Thevenin resistance (sources off)

R_th = 8 + (5 ∥ 2) = 8 + 1.4286 = 9.429 Ω

Step 3: Current in the 10 Ω

I = |V_th|/(R_th + 10) = 6.286/(9.429 + 10) = 6.286/19.429 = 0.3235 A

The current flows upwards through the 10 Ω (from the bottom rail towards the 8 Ω side), because the open terminal is below the rail potential.

Answer: I₁₀Ω ≈ 0.324 A

Assumption: battery polarities are read from the plate lengths in the figure.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Basic Electrical Engineering chapter 3 questions →