Circuit Analysis

Dear All,

I am just reading one book for circuit analysis and struck on one point. I need to find out voltage drop across two resistors individually. I am attaching the pic with my question please guide me how i can find the voltage drop across resistor one and then across resistor-2 as i know total voltage drop will be equal to the applied voltage to the circuit.

Hi ASAD,
'ea' is 2v because it is for voltage source .
and you can transform to Norton. then you have 3 shunt element .2 resistor and 1 current source with 3A. 'eb' is voltage for each resistor equal to 3/2=1.5

Here's how I would solve it:

By *Kirchhoff's second law* -- all of the voltages in the loop will add to zero:

*2V = Iab(Rab) + Ibc(Rbc)* [since the voltage across Rbc is the same as the voltage across the 1A current source]

And, by *Kirchhoff's first law* -- the sum of the currents flowing into a node is equal to the sum of the currents flowing out of the same node:

*Iab = Ibc + 1A*

Solve for *Ibc*:

*-Ibc = Iab/1-*

Thus by algebraic substitution:

-*2 = Iab(Rab) + Iab(Rbc)/1*-
-*2 = Iab[Rab + Rbc/1]*-
-*2 = Iab[1 + 1/1] = 2Iab*-
-*Iab = 1*-

Ohms law states that the voltage across a resistor is equivalent to the current flowing through that resistor multiplied by the resistor's value:

-*Vab = Iab(Rab) = 1(1) = 1V*-

At this point we can conclude that the voltage across *Rbc* is also *1V* -- [*Kirchhoff's second law*] but, let's see if the math corroborates with that:

-*Ibc = Iab/1 = 1/1 = 1*-
-*Vbc = Ibc(Rbc) = 1/1 = 1V* ✔-

OK, made a mistake -- I started wondering how there could be a 1A source AND 1A flowing through Rbc and yet also 1A flowing through Rab -- a violation of *Kirchhoff’s first law*.

Then I found my mistake (where I "solved" for *Ibc*):

-*Iab = Ibc + 1*-
-*Ibc = Iab - 1*-

*∴*

-*2 = Iab(Rab) + Rbc[Iab - 1]*-
-*2 = Iab(Rab) + Rbc(Iab) - 1(Rbc)*-
-*2 = Iab(1) + Iab(1) - 1(1)*-
-*2 = 2Iab - 1*-
-*3 = 2Iab*-
-*Iab = 1.5*-

-*Vab = Iab(Rab) = 1.5(1) = 1.5V*-

-*Ibc = Iab - 1 = 1.5 - 1 = 0.5A*-
-*Vbc = Ibc(Rbc) = 0.5(1) = 0.5V*-

(2-Eb)/1 = Eb/1 + 1

Eb = 1/2 V

Sajed:

If the voltage across each resistor is *1.5V* then *Kirchhoff’s second law* is violated:

*2V = 1.5V/1Ω + 1.5V/1Ω*
*2V ≠ 3V*

Also, it can be shown that there is no way to satisfy *Kirchhoff’s first law* when both resistors have *1.5v* across them:

*-1.5/Rab - 1.5/Rbc + 1A = -1.5/1 - 1.5/1 + 1 = -2 ≠ 0*
*1.5/Rab - 1.5/Rbc + 1A = 1 ≠ 0*
*-1.5/Rab + 1.5/Rbc + 1A = 1 ≠ 0*

Whereas, this does satisfy *Kirchhoff’s first law*:

*-1.5/Rab + 0.5/Rbc +1 = 0 ✔*

And the second law:

*2 - Iab(Rab) - Ibc(Rbc) = 2 - (-1.5)(1) - (0.5)(1) = 0 ✔*

Please guide me sir steve

You're welcome

Thank you soo much Steve sir

Dear Steve,
No, I change circuit to "Norton" model and solve it. it mean voltage source and series resistor change to current source with shunt resistor. at this condition vbc is for each shunt resistor.
but you solve it without change circuit.
thank you.
"please see this image, in norton mode(right) 2 resistor is shunt and have 1.5 volt()":http://sajedrakhshani.persiangig.com/Norton%20model.png

Holy crustacean! Sajed is right (at least about the voltage across *Rbc*)!

I made the assumption that *Iab = Ibc + 1*, thinking that if my assumption was incorrect, I would get nonsensical numbers. Apparently not! Based on Sajed's simulation (and one that I also did), the truth is:

*Ibc = Iab + 1A*

So, if

*2V = Iab(Rab) + Ibc(Rbc)*

Then

*2 = Iab(Rab) + (Iab + 1)Rbc*
*2 = Iab(1) + (Iab + 1)1*
*2 = Iab + Iab + 1*
*2 = 2(Iab) + 1*
*Iab = 0.5*

*∴*

*Vab = Rab(Iab) = (1)(0.5) = 0.5V* [*Not 1.5V*]
*Vbc = 2 - Vab = 1.5V!*

So, Sajad, you're correct that the voltage across *Rbc* is *1.5V*, but how can the voltage across *Rab* also be *1.5V*? That would add up to *3V* (and the simulator agrees with me).

What I've learned from this is, after 30 (or so) years of non-use, my DC analysis skills are rusty, so I will remain mute for further questions like this!

Well, not so fast, young man--I completely botched this! Hopefully, this wasn't a homework problem (or worse, a test question)

Hi Steve,
Rab is not 1.5, I solve it with "Norton equivalent circuit" and in Norton equivalent circuit series voltage source must change into shunt current source,in this equivalent circuit *R'ab* is not equal to *Rab*, but in original circuit Rab is 0.5.
best wishes for you.

Clearly the correct way to do it -- well done!

Ohm's Law. I = V/R

Rearrange it to be V = IR.

The current across the resistor multiplied by its resistance gives the voltage across it. Simple.

Didn't see the current source. That definitely makes it more complicated than I thought. This is the way I see it:

Remove the resistor in parallel with the current source. Now, the current source wants to push three volts above ground. Add the resistor back in to get a voltage divider. It divides the voltage pushed by the current source into two. Now, the voltage across the resistor in parallel with the current source is seeing one and a half volts, leaving half a volt for the other resistor.

I also didn't see the most recent post date. I apologize for reviving an old post if it was solved.