Skip to main content

Thevenin: 4 kΩ resistor, 12 V, 10 mA

Short answer

Thevenin's theorem: Vth is open-circuit voltage, Rth is resistance with sources zeroed. Here Vth = 8V − 20V = −12V, Rth = (3k∥6k)+2k = 4kΩ. Current through 4kΩ is 1.5 mA from B to A.

State Thevenin's theorem and use it to find the current through the 4 kΩ resistor. [Figure: 12 V source feeding a 3 kΩ series resistor, then 6 kΩ and 2 kΩ resistors in parallel branches, then a 4 kΩ series resistor to a 10 mA current source in the last branch.]

Statement: Thevenin's theorem says that any linear two-terminal network containing independent sources and resistors can be replaced by one voltage source Vth in series with one resistance Rth. Vth is the open-circuit voltage across the terminals and Rth is the resistance seen at the terminals with voltage sources shorted and current sources opened. The load current is then I = Vth/(Rth + RL).

Assumed circuit (the figure was unclear): the 12 V source feeds node A through 3 kΩ. A 6 kΩ resistor is connected from A to ground. The 4 kΩ load connects A to node B. At node B, the 2 kΩ resistor goes to ground, and the 10 mA source (pointing into B) is also connected to ground. Load: 4 kΩ between A and B.

Step 1: remove the 4 kΩ load, find Vth = VA − VB

  • Left side (A): VA = 12 × 6/(3 + 6) = 8 V
  • Right side (B): the 10 mA flows through 2 kΩ: VB = 10 mA × 2 kΩ = 20 V

Vth = VA − VB = 8 − 20 = −12 V

Step 2: find Rth

Short the 12 V source and open the 10 mA source.

  • Left: 3 kΩ ∥ 6 kΩ = 2 kΩ
  • Right: 2 kΩ

Rth = 2 kΩ + 2 kΩ = 4 kΩ

Step 3: load current

I = Vth/(Rth + RL) = −12 V/(4 kΩ + 4 kΩ) = −1.5 mA

Answer: I = 1.5 mA through the 4 kΩ resistor, flowing from B to A (opposite to the A to B reference), under the stated assumption.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Basic Electronics Engineering chapter 1 questions →