Thevenin theorem, current in 13Ω

Reading the figure

• Left branch: 200 V battery (+ on top) joins node N₁ (bottom) to the top node T (V_T = V_N1 + 200).
• 50 V battery (+ on top) below N₁: its + end is at N₁ and its − end is terminal a.
• 15 Ω from T to the middle node N₂; 20 Ω from N₁ to N₂; 10 Ω from N₂ to the right node N₃.
• 180 V battery (+ on the T side) between T and N₃ (V_N3 = V_T − 180).
• The load 13 Ω is between a and b, and b is wired to N₃.

Take N₁ as reference (0 V).

Step 1: Thevenin voltage (13 Ω removed)

• V_T = 200 V
• V_N3 = V_b = 200 − 180 = 20 V
• No current flows in the a-b path, so V_a = −50 V (50 V battery, − terminal)

V_b − V_a = 20 − (−50) = 70 V

V_th = 70 V (b is positive with respect to a).

(The resistors do not change V_b or V_a: V_N2 settles at 70.8 V by KCL at N₂, but it does not enter V_ab.)

Step 2: Thevenin resistance (sources off)

With the three batteries replaced by short circuits, a is joined to N₁, N₁ to T, T to N₃ and N₃ to b by wires. So a and b become the same point:

R_th = 0 Ω

Step 3: Current in the 13 Ω

I = V_th/(R_th + 13) = 70/13 = 5.385 A

Direction: V_b > V_a, so the current flows through the 13 Ω from b to a (right to left).

Answer: I = 5.38 A, from b to a