Electrical Engineering - Circuit Theorems and Conversions - Discussion

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.

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.

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).

Simply the answer will be the difference of two current sources.

Why because the current from two sources enters the branch in opposite directions. So current through that branch will be I = 0.2-0.04 = 0.16 A.

What ever is the resistance placed in that branch the current through that branch will be 0.16 A.

Is1 & Is2 are independent time-invariant current sources. Is1 divides into two branches.
That's why the current through 120 ohms branch is different between Is1 & Is2. The current will not change either 120ohms is changed to 68ohms since it has an independent current source.

When source 2 is active then both r1 and r2 are in series so same current i.e 0.04A.
When source 1 is active then only r1 is active because r2 is open circuited so r1 gets 0.2A.
Hence net current through r1 = 0.2 - 0.04 = 0.16.

If resistance decreses current increases at node current flow 0.2 amps curent divide each branch then resistance is low curent flow high so 0.16

0.16 + 0.02 = 0.2 amps.

Only IS1 is active (And IS2 current travel to the ground).

Use current divider formula I2 = Im (R1/(R1+R2)).

= 0.2*(68/ (68+68)).

= 0.1 A.

Near is 0.16 A.

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.

In each branch there should be equal currents.... means 0.1amp. Branch r2 already hve current source of .04 amp.

So current in ri is 0.1+.06 = 0.16amp