Question

Using Kirchhoff's rules, find the currents I1, I2, and I3 in the circuit shown where R₁ = 1.2 Ω, R₂ = 2.8 Ω, and R₃ = 6.3 Ω.

Answer 1

Using kirchoff's Junction law:

I1 + I2 = I3

Using kirchoff's Voltage law in upper loop:

10 + I2*R1 - I1*R2 = 0

10 = 2.8*I1 - 1.2*I2

Using kirchoff's voltage law in lower loop:

R2*I1 + R3*I3 = 0

I3 = -R2*I1/R3

I3 = -2.8*I1/6.3

I3 = -0.444*I1

Now we have three equations and three unknowns:

I1 + I2 = I3

10 = 2.8*I1 - 1.2*I2

I3 = -0.444*I1

Using 3rd equation:

I2 = I3 - I1 = -0.444*I1 - I1 = -1.444*I1

Now using 1nd equation:

10 = 2.8*I1 - 1.2*(-1.444*I1)

I1 = 10/(2.8 + 1.2*1.444)

I1 = 2.21 Amp

I2 = -1.444*2.21 = -3.19 Amp

I3 = -0.444*2.21 = -0.98 Amp

(-ve sign means direction of current will be in opposite direction then shown, if you need magnitude then use positive value, otherwise use negative)