## Related Posts

1. Samuel says:

Hope you will solve more solutions

2. Harsha says:

Solve problems on all theorems ?

3. P. Pravallika says:

Very good

4. Kwashie Andoh says:

In step iii,
Since we need current through 15 ohm resistor, shouldn’t it be:
I3= 10*15/(15+40) ?

1. Abragam Siyon Sing says:

No. that’s the wrong way. Since it is a parallel circuit, the current division rule is applied. Kindly go through the rule to get a clear idea.

5. BLESSY says:

very good notes.Easily understandable.Thank you

1. Abragam Siyon Sing says:

Glad to know that the contents are understandable. Thank you.

6. Raj says:

Sir actually i want to ask u a numerical question related to superposition theorem how i can sen u the question

7. Cherry says:

Sir, in the second problem to actually verify superpositon theorem v total should also be calculated when all the sources are active which you didn’t mention there will you please explain the process involved there in detail?

1. Abragam Siyon Sing says:

The superposition theorem is used not to verify but to solve complex electric circuits. To verify this theorem, you can either simulate the circuit in any simulation software or you can do it practically.
In the superposition theorem, the current or voltage due to each individual source is determined and then they are added to find the resultant(total) current or voltage across the given load. The same is followed for both the problems.
kindly check the above steps to solve.

1. Gilbert Shadrack says:

Am so glad because my problems in this area are now fully solved.

8. rio says:

for problem 2, step (iv) can you please explain how/where the low resistance path works and why the 15 ohm resistor has a negligible resistance.
thanks

1. Abragam Siyon Sing says:

in step (iv), for the 5A source, there are two paths for the flow of current. One path is through 15Ω resistance and the other path is the short-circuited path, which is across the current source.
In an electrical circuit, a short circuit will have very low resistance.
Since there is very low resistance in this path, all the current from the source rushes through the short circuit path, making a negligible amount of current to flow through 15Ω resistance.
That is the reason, I4 =0.
And in your question, you have asked, “why the 15 ohm resistor has a negligible resistance.”
15Ω resistor will never have negligible resistance. It has its own resistance of 15Ω.

9. Frelie C. Poblete says:

In part iv, to find for V4, you said the other current source is open-circuited but on the re-drawing of the circuit it is short-circuited.

1. Abragam Siyon Sing says:

Yeah, you are right. It was corrected now and thanks for showing the mistake.

2. El Salif says:

I don’t also understand why v4 in problem 2 is completely zero.
I try using Ohms law for it, it was 0.060v.
Can you help explain to me more better.

1. Abragam Siyon Sing says:

Could you please tell me how you applied Ohm’s law?

3. Innocent says:

I don’t know how I can easily understand part iv, the reason for completely ignoring 5A, does it mean it’s not contributing any voltage to resistor 15, if it does then why not considering it.
Secondly, I am asking, in part iii, when we connect to 10A, resistors 15 and 40 are considered to be in parallel, what about when on IV, when connected to 5A, are they in parallel or series?