Thevenin: 90, 50, 130 ohm circuit

Thevenin's theorem: any linear two-terminal network containing sources and resistors can be replaced by a single voltage source Vth in series with a single resistance Rth. Vth is the open-circuit voltage across the two terminals and Rth is the resistance seen from the terminals with all independent sources made zero (voltage sources shorted, current sources opened).

Principle shown by the problem: the load R3 is removed, the rest of the circuit is replaced by Vth and Rth, and then the load current is found from one simple series loop.

Assumption: the low-resolution figure does not show the polarity of E2, so both batteries are taken with their positive terminals toward the top node (both help to drive current through R3). E1 = 100 V, R1 = 90 Ω, E2 = 60 V, R2 = 50 Ω, R3 = 130 Ω. The three branches are in parallel across the same two nodes.

Step 1: remove R3 and find Vth

With R3 removed, only a circulating current flows through E1, R1, R2 and E2:

I' = (E1 − E2)/(R1 + R2) = (100 − 60)/(90 + 50) = 0.2857 A

Vth = E1 − I'R1 = 100 − 0.2857 × 90 = 74.29 V

(Check: 60 + 0.2857 × 50 = 74.29 V.)

Step 2: find Rth

Short both batteries. R1 and R2 are then in parallel as seen from the open terminals:

Rth = (90 × 50)/(90 + 50) = 4500/140 = 32.14 Ω

Step 3: Thevenin equivalent

     Rth = 32.14 ohm
 +----/\/\/\----+
 |              |
 Vth = 74.29 V  R3 = 130 ohm
 |              |
 +--------------+

I = Vth/(Rth + R3) = 74.29/(32.14 + 130) = 74.29/162.14

Answer: I = 0.458 A (about 458 mA) through R3 (130 Ω)