Skip to main content

Mesh analysis with 3 A source

Short answer

Right mesh current is forced to -3 A by the 3 A source. KVL on left mesh: 12 = 7I₁ – 2(-3) gives I₁ = 0.857 A. Thus mesh currents are I₁ = 0.857 A (clockwise), I₂ = -3 A (3 A anticlockwise).

Determine the mesh current using Mesh Analysis for the network shown in figure below. [Figure: left mesh has 5 Ω in the top branch, a 12 V source on the left side, and 2 Ω in the middle vertical branch; top branch carries a 3 A current source (arrow pointing left) across the top; right mesh has 6 Ω in its top branch, a 6 V source on the right side, and 1 Ω in the bottom branch shared with the bottom rail.]

Assumptions (from the figure): the 3 A current source is in the top branch of the right mesh in series with the 6 Ω resistor, and it forces 3 A to flow leftwards there. Mesh currents I1I_1 (left) and I2I_2 (right) are both taken clockwise. The 12 V source aids I1I_1 and the 6 V source opposes I2I_2.

Step 1: Right mesh (current source in the branch). Clockwise I2I_2 flows to the right along the top branch, but the source forces 3 A to the left, so

I2=−3 AI_2=-3\ \text{A}

A current source in a mesh needs no KVL equation for that mesh; its current fixes the mesh current directly.

Step 2: KVL in the left mesh.

12=5I1+2(I1−I2)12=7I1−2I27I1=12+2(−3)=6I1=67=0.857 A\begin{aligned} 12 &= 5I_1 + 2(I_1-I_2)\\ 12 &= 7I_1 - 2I_2\\ 7I_1 &= 12 + 2(-3) = 6\\ I_1 &= \frac{6}{7} = 0.857\ \text{A} \end{aligned}

Step 3: Current in the 2 Ω shared branch.

I2Ω=I1−I2=0.857−(−3)=3.857 AI_{2\Omega}=I_1-I_2 = 0.857-(-3)=3.857\ \text{A}

Answer: I1=0.857I_1 = 0.857 A (clockwise), I2=−3I_2 = -3 A (i.e. 3 A anticlockwise, fixed by the source); current in the 2 Ω resistor = 3.857 A.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Basic Electrical and Electronics Engineering chapter 1 questions →