Electrical Engineering - Circuit Theorems and Conversions - Discussion

Condition given that R1=68 and R2=68.

Apply KCL at central node.

So current flows equally to both the resistors.

Putting superposition theorem into action.

Current due to Is1(Is2 removed) in replaced 68 ohms resistance alone = 0.2 A.
Current due to Is2(Is1 removed) in replaced 68 ohms resistance alone = 0.04 A.

Adding in reference with direction = 0.2-0.04 = 0.16 A.
(As current in the replaced 68 ohm resistance is opposite in both cases).

one should note that the value of current flowing in a branch is unique and hence the current in the branch containing IS2 is 0.04.

Now the current 0.2A divides at the central node and the remaining current (0.2-0.04=0.16A) flows through 120ohm branch.

The point to be noted is, even if the resistor value is changed from 120 to 68 ohms,the current through the branch remains the same at 0.16A.

This is supported by the fact that the current delivered by an ideal source is independent of the load or voltage across it.

As both are current sources the current won't change at node hence.

Current in 120 ohm resistor = current in 68 ohm resistor.

I = 0.2-0.04 = 0.16A.

Apply superposition theorem; Acc.to this in first case short circuit the current source IS2 hence we have current across R2 = I*R1/R1+R2 = 0.2A*120/188 = 0.12 say (I1).

Now; in second case short circuit the current source IS1 hence we have current across R2 = 0.04 say (I2) [because current flow in least resistance path]

Hence total current across R2 = I1+I2 = 0.12+0.04 = 0.16A answer.

By Using current source theorems 0.2-0.04 = 0.16.

Applying kcl in the circuit.

We will get,
I1 + 0.04 = 0.2.
We get I1 = 0.16.

As per KCL,
Entering current = leaving current.

I1 = I2.
I1-I2 = 0.
0.2-0.04.

I think 0.16 on -ve direction.

Simple ans:

R1 = 120, R2 = 68, IS1 = 0.2, IS2 = 0.04.

Change to R1 = 68.

So IS1 changed 0.04 then,

IS1 x IS2 = 0.2 x 0.2 = 0.16.