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.

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).

Condition given that R1=68 and R2=68.

Apply KCL at central node.

So current flows equally to both the resistors.

Apply kcl v1/68=2-.04.

v1/68 = 0.16.

v1 = 0.16*68.

v1 = i1*r1.

0.16*68 = i1*68.

i1 = 0.16A.

By kirchoff's current law.

I +0.2=0.04.

I=0.04-0.2 = -0.16A.

v=.04*68=2.72 (fixed)
so
v at new 68 = 2.72 (parallel)
i=2.72\68 = .04 a

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.

Apply kcl at centre node
0.2-0.04=1.6

I1=.2A,when I1 active
I2=.04A,when I2 active
Final current=.2-.04=.16A(ans)

Jyoti can you explane how your gating 0.06 in last step?