Electrical Engineering - Circuit Theorems and Conversions - Discussion

From super position theorem

case 1: using only Vs1 and shorting Vs2.

Rt=R1+(R2||R3)

Rt=180 ohm;

It=Vs1/Rt

It=12/180

It=0.067

I(R2)=(R2||R3)*It/R3

I(R2)=0.0333A=33.4mA

case 2: using only Vs2 and shorting Vs1.

Rt=R3+(R1||R2)

Rt=180 ohm

It=Vs2/Rt

It=6/180=0.0333

I(R2)=(R2||R1)*It/R1

I(R2)=16.7mA

Adding case1 I(R2)+ case2 I(R2), because both currents are flowing in same direction.

We get 33.3mA+16.7mA=50mA

I1 = Vs1/R1.
= 12/120 = 0.1.

I2 = Vs2/R3 .
= 6/120 = 0.05.

I = I1 + I2 .
= 0.1 + 0.05 = 0.15.

I(R2) = I * R2/(R1 + R2 + R2).
= 0.15 * 120/360.
= 50mA.

In this question also apply the millman's theorem and it was very easy and fast technique.

Voltage for R2 = Vs1-Vs2.
12-6 = 6v.
Try the different options V/R.
6/50mA = 120.

I(R2) = V/R2.
= 6/120 = 1/20 = 0.05A = 50mA.

Vs1/r1+r2.
i.e.12/120 + 120.
= 50 * 10^-3.