Electrical Engineering - Circuit Theorems and Conversions - Discussion

Apply SPT,

For S1 active R1 = 0.2A(S2 = 0).
FOR S2 active r1 = -0.04A(s1 = 0).

Then I(total) = 0.2+(-0.04) = 0.16.

If in the question current source are given then one thing I want to say that current source does not allow current through itself. So option B is correct.

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.

I think 0.16 on -ve direction.

As per KCL,
Entering current = leaving current.

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

Applying kcl in the circuit.

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

By Using current source theorems 0.2-0.04 = 0.16.

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.

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.

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.