Kirchhoff law currents and source power

Reading the figure

Node P (top middle) joins the 3 Ω (from the left branch), the 7 Ω and the vertical 5 Ω. Node M is the bottom of that 5 Ω; from M, one 5 Ω goes to the ground rail. Node S is the right-hand node, joined to M by a 5 Ω.

• Left branch: 10 V battery (+ on top) with 2 Ω and 3 Ω into P. Current I₁ flows into P.
• Right top: 7 Ω from node Q to P. Q connects through 3 Ω and a 5 V battery (+ on top) to S. Current I₂ flows from Q to P.
• S to M: 5 Ω carrying I₃.
• S to ground: 5 Ω with a 5 V battery (+ on top).

Polarities read from the plate lengths in the scan.

Currents by KCL

• 5 Ω from P to M: I₁ + I₂
• 5 Ω from M to ground: I₁ + I₂ + I₃
• Lower right branch (towards S): I₂ + I₃

KVL equations

Left loop (10 V, 2 + 3 Ω, 5 Ω, 5 Ω): 10 = 5 I₁ + 5 (I₁ + I₂) + 5 (I₁ + I₂ + I₃) ⇒ 3 I₁ + 2 I₂ + I₃ = 2 ... (1)

Middle loop (P, Q, 3 Ω, 5 V, S, M): 7 I₂ + 3 I₂ − 5 − 5 I₃ + 5 (I₁ + I₂) = 0 ⇒ I₁ + 3 I₂ − I₃ = 1 ... (2)

Lower loop (5 V, 5 Ω, S–M 5 Ω, 5 Ω to ground): 5 = 5 (I₂ + I₃) + 5 I₃ + 5 (I₁ + I₂ + I₃) ⇒ I₁ + 2 I₂ + 3 I₃ = 1 ... (3)

Solution

From (2) and (3): subtracting gives I₂ = 4 I₃. Then (1) and (2) give I₃ = 1/24.

I₁ = 13/24 = 0.542 A, I₂ = 1/6 = 0.167 A, I₃ = 1/24 = 0.042 A

Power output of each source

All three sources deliver power; total = 7.29 W.