Skip to main content

Mesh analysis: find Vs for 6.76 A in 3 Ω

Short answer

Mesh currents: I2 = 6.76 A, I3 = 10 A, 3 A source gives I1 = I2 + 3 = 9.76 A. Supermesh KVL yields Vs = 5·9.76 + 5·6.76 + 10·(6.76 − 10) + 1·(9.76 − 10) = 49.96 V ≈ 50 V.

Using mesh analysis, determine the value of Vs so that the current through the 3 Ω resistor is 6.76 Amp as shown in the following figure. [Figure: voltage source Vs (+ at top) in series with a 5 Ω resistor on the left side, feeding a top-left node A. From A: a 3 Ω resistor (carrying the 6.76 A, direction left to right) and then a 2 Ω resistor along the top wire to the top-right node C; a 3 A current source (arrow pointing down-right) from A diagonally to the middle node M; from C a 10 Ω resistor diagonally to M; a 10 A current source (arrow pointing down) from C to the bottom wire; a 1 Ω resistor from M to the bottom wire. Exact connection of the 3 Ω and 2 Ω junction is not clear in the scan.]

Assumptions (figure unclear in the scan): Vs with the 5 Ω resistor feeds node A. The 3 Ω and 2 Ω resistors are in series from A to C and both carry the 6.76 A. The 3 A source goes from A to M, a 10 Ω resistor from C to M, a 10 A source from C to ground, and a 1 Ω resistor from M to ground.

Mesh currents (clockwise)

  • I₁: left mesh (Vs, 5 Ω, 3 A source, 1 Ω)
  • I₂: top mesh (3 Ω, 2 Ω, 10 Ω, 3 A source), so I₂ = 6.76 A
  • I₃: right mesh (10 Ω, 10 A source, 1 Ω), so I₃ = 10 A (the 10 A source is only in this mesh)

Current source constraint

The 3 A source is shared by I₁ and I₂ (arrow A to M):

I₁ − I₂ = 3  →  I₁ = 6.76 + 3 = 9.76 A

Supermesh of I₁ and I₂ (excluding the 3 A source)

KVL around Vs, 5 Ω, 3 Ω, 2 Ω, 10 Ω, 1 Ω:

Vs = 5·I₁ + (3 + 2)·I₂ + 10(I₂ − I₃) + 1(I₁ − I₃)
   = 5(9.76) + 5(6.76) + 10(6.76 − 10) + (9.76 − 10)
   = 48.8 + 33.8 − 32.4 − 0.24
   = 49.96 V

Answer: Vs ≈ 50 V (49.96 V).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electrical Circuit Theory chapter 1 questions →