Let R1=10Ω, R2=20Ω, R3=40Ω, V1=10V, V2=20V.

• We establish the senses of the \$I_{1}\$ and \$I_{2}\$ mesh currents, taking into account the sources and motors of the circuit.
• We need two equations to be able to clear both variables. To do this, we apply the mesh rule twice.

Left mesh: 10 = 40(I2-I1)-10(I1) = -50I1 + 40I2
Right mesh: 20 = 20(I2)+40(I2-I1) = -40I1 + 60I2

Solving the equations:

• \$I_{1}\$ ≈ 0.14286A

• \$I_{2}\$ ≈ 0.42857A

The intensity that passes through each resistance:

• \$R_{1}=10Ω\ \rightarrow I_{1}\$ ≈ 0.14286A
• \$R_{2}=20Ω\ \rightarrow I_{2}\$ ≈ 0.42857A
• \$R_{3}=40Ω\ \rightarrow I_{3}\$ = \$I_{2}\$ - \$I_{1}\$≈ 0.42857A - 0.14286A=0.28571A

I'm not sure how to go about this, but the answers are below

I'm not sure how to go about this, but the answers are below

Let R1=10Ω, R2=20Ω, R3=40Ω, V1=10V, V2=20V.

• We establish the senses of the \$I_{1}\$ and \$I_{2}\$ mesh currents, taking into account the sources and motors of the circuit.
• We need two equations to be able to clear both variables. To do this, we apply the mesh rule twice.

Left mesh: 10 = 40(I2-I1)-10(I1) = -50I1 + 40I2
Right mesh: 20 = 20(I2)+40(I2-I1) = -40I1 + 60I2

Solving the equations:

• \$I_{1}\$ ≈ 0.14286A

• \$I_{2}\$ ≈ 0.42857A

The intensity that passes through each resistance:

• \$R_{1}=10Ω\ \rightarrow I_{1}\$ ≈ 0.14286A
• \$R_{2}=20Ω\ \rightarrow I_{2}\$ ≈ 0.42857A
• \$R_{3}=40Ω\ \rightarrow I_{3}\$ = \$I_{2}\$ - \$I_{1}\$≈ 0.42857A - 0.14286A=0.28571A