How to Calculate Voltage Drop Across Two Series Resistors in a Circuit?

162 16

#1 21670071 11 Oct 2013 12:04

ASAD ALI ASAD ALI

Anonymous

Post #1
21670071 11 Oct 2013 12:04

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.
ADVERTISEMENT
#2 21670072 11 Oct 2013 12:41

Sajed Rakhshani Sajed Rakhshani

Anonymous

Post #2
21670072 11 Oct 2013 12:41

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
#3 21670073 11 Oct 2013 12:48

Steve Lawson Steve Lawson

Anonymous

Post #3
21670073 11 Oct 2013 12:48

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* ✔-
ADVERTISEMENT
#4 21670074 11 Oct 2013 13:23

Steve Lawson Steve Lawson

Anonymous

Post #4
21670074 11 Oct 2013 13:23

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*-
ADVERTISEMENT
#5 21670075 11 Oct 2013 13:27

DAVID CUTHBERT DAVID CUTHBERT

Anonymous

Post #5
21670075 11 Oct 2013 13:27

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

Eb = 1/2 V
#6 21670076 11 Oct 2013 13:39

Steve Lawson Steve Lawson

Anonymous

Post #6
21670076 11 Oct 2013 13:39

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 ✔*
#7 21670077 11 Oct 2013 13:59

ASAD ALI ASAD ALI

Anonymous

Post #7
21670077 11 Oct 2013 13:59

Please guide me sir steve
#8 21670078 11 Oct 2013 14:01

Steve Lawson Steve Lawson

Anonymous

Post #8
21670078 11 Oct 2013 14:01

You're welcome
#9 21670079 11 Oct 2013 14:07

ASAD ALI ASAD ALI

Anonymous

Post #9
21670079 11 Oct 2013 14:07

Thank you soo much Steve sir
#10 21670080 11 Oct 2013 23:27

Sajed Rakhshani Sajed Rakhshani

Anonymous

Post #10
21670080 11 Oct 2013 23:27

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
#11 21670081 12 Oct 2013 09:04

Steve Lawson Steve Lawson

Anonymous

Post #11
21670081 12 Oct 2013 09:04

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!
#12 21670082 12 Oct 2013 09:06

Steve Lawson Steve Lawson

Anonymous

Post #12
21670082 12 Oct 2013 09:06

Well, not so fast, young man--I completely botched this! Hopefully, this wasn't a homework problem (or worse, a test question)
#13 21670083 13 Oct 2013 04:15

Sajed Rakhshani Sajed Rakhshani

Anonymous

Post #13
21670083 13 Oct 2013 04:15

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.
#14 21670084 13 Oct 2013 09:56

Steve Lawson Steve Lawson

Anonymous

Post #14
21670084 13 Oct 2013 09:56

Clearly the correct way to do it -- well done!
ADVERTISEMENT
#15 21670085 04 Nov 2013 02:25

Cody Gass Cody Gass

Anonymous

Post #15
21670085 04 Nov 2013 02:25

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.
#16 21670086 04 Nov 2013 02:40

Cody Gass Cody Gass

Anonymous

Post #16
21670086 04 Nov 2013 02:40

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.
#17 21670087 04 Nov 2013 02:40

Cody Gass Cody Gass

Anonymous

Post #17
21670087 04 Nov 2013 02:40

I also didn't see the most recent post date. I apologize for reviving an old post if it was solved.
Create an account, log in here. You will receive points by participating in discussions.
Join this discussion.

Install Elektroda application

Didn't find an answer? Ask Artificial Intelligence

*I agree to send the question to OpenAI, Anthropic PBC, Perplexity AI, Inc., Kagi Inc., Google LLC - owners of language models in order to prepare the best response. The companies may monitor and log information entered into the form.

*I agree to publicly display my question and answer. The question and answer will be publicly available to everyone. The process may take a few minutes. Upon completion, you will be redirected to the page with the answer.

Wait...(2min)

Topic summary

✨ The discussion addresses calculating voltage drops across two series resistors in a circuit with a voltage source and a current source. The problem involves applying Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL) to determine individual voltage drops. Initial attempts incorrectly assumed current relationships, leading to inconsistent voltage values violating Kirchhoff’s laws. Correct analysis shows that the current through the first resistor (Rab) and the second resistor (Rbc) differ due to the presence of a 1A current source in parallel. Using the Norton equivalent circuit transformation (converting the voltage source and series resistor into a current source with parallel resistors) clarifies the problem. The final correct solution finds the current through Rab as 0.5A, resulting in a voltage drop of 0.5V, and the voltage drop across Rbc as 1.5V, satisfying both KCL and KVL. Ohm’s Law (V=IR) is fundamental in these calculations. The discussion highlights the importance of correctly modeling the circuit and carefully applying circuit laws to solve for voltage drops in complex circuits involving current sources.

Home page
/
Forum
/
Circuit Design
/
EEWeb Topics
/
How to Calculate Voltage Drop Across Two Series Resistors in a Circuit?

FAQ

TL;DR: For a 2 V source with two 1 Ω resistors and a 1 A current source between them, "Vab = 0.5 V; Vbc = 1.5 V." Solve using KCL (Ibc = Iab + 1) and KVL (2 = Vab + Vbc). [Elektroda, Steve Lawson, post #21670081]

Why it matters: This FAQ helps students and hobbyists quickly calculate individual voltage drops in mixed source series circuits without sign errors.

Quick Facts

Solved values here: Vab = 0.5 V, Vbc = 1.5 V; Iab = 0.5 A, Ibc = 1.5 A. [Elektroda, Steve Lawson, post #21670081]
Correct node relation: Ibc = Iab + 1 A (not Iab = Ibc + 1 A). [Elektroda, Steve Lawson, post #21670081]
Norton tip: Convert the series voltage source + resistor to a parallel current source + resistor; parallel legs share the same voltage. [Elektroda, Sajed Rakhshani, post #21670080]
Law check that catches mistakes: 2 − Iab·1 − Ibc·1 = 0 and KCL balance at the node. [Elektroda, Steve Lawson, post #21670076]
Ohm’s law refresher: V = I·R; use it after solving currents to get drops. [Elektroda, Cody Gass, #21670085

How do I find the voltage drop across each of the two 1 Ω resistors in this mixed-source loop?

Write KCL: Ibc = Iab + 1. Write KVL: 2 = Iab·1 + Ibc·1. Solving gives Iab = 0.5 A and Ibc = 1.5 A. Then Vab = 0.5 V and Vbc = 1.5 V. [Elektroda, Steve Lawson, post #21670081]

What equations should I write first (KCL/KVL) to avoid sign mistakes?

Use KCL at the middle node: Ibc = Iab + 1. Then apply KVL around the loop: 2 = Vab + Vbc = Iab·1 + Ibc·1. Solve for currents, then voltages. “KCL plus KVL nails it.” [Elektroda, Steve Lawson, post #21670081]

Why did someone get 1.5 V across both resistors, and why is that wrong?

That came from a sign error in KCL. If both drops were 1.5 V, KVL would give 3 V, contradicting the 2 V source. The KCL sum at the node would also fail to balance. [Elektroda, Steve Lawson, post #21670076]

What are the actual currents through each resistor here?

From KCL and KVL, Iab = 0.5 A through the left 1 Ω resistor, and Ibc = 1.5 A through the right 1 Ω resistor. That’s a 3× current difference due to the 1 A current source. [Elektroda, Steve Lawson, post #21670081]

How can Norton’s theorem simplify this problem?

Convert the series voltage source and its resistor to a Norton current source in parallel with a resistor. In that equivalent, the two resistors are in parallel and share the same node voltage, clarifying the 1.5 V on the right branch. [Elektroda, Sajed Rakhshani, post #21670080]

What is a Norton equivalent circuit?

It’s an equivalent representation of any linear two-terminal network as a current source in parallel with a resistor. Here, converting helps visualize parallel elements sharing the same voltage. [Elektroda, Sajed Rakhshani, post #21670083]

Can you give me a quick 3-step method to solve similar circuits?

Write KCL at the critical node(s) to relate branch currents.
Write KVL around the loop(s) to relate voltages and currents.
Use Ohm’s law to compute individual drops and verify with KCL/KVL. [Elektroda, Steve Lawson, post #21670073]

How do I compute power in each resistor for this case?

Use the solved values: Vab = 0.5 V and Vbc = 1.5 V across 1 Ω each. Power is Pab ≈ 0.25 W and Pbc ≈ 2.25 W. Higher power on the right branch reflects the larger 1.5 A current. [Elektroda, Steve Lawson, post #21670081]

What common sign convention pitfall should I watch for?

Be consistent with current directions when writing KCL. Swapping the relation to Iab = Ibc + 1 leads to incorrect voltages and KVL failure. “Mind the arrows; math follows.” [Elektroda, Steve Lawson, post #21670076]

If the resistor values change, how do I adapt the solution?

Keep the same equations: KCL at the node and KVL around the loop. Replace 1 Ω with your actual Ra and Rb, solve currents, then use V = I·R to get each drop. [Elektroda, Steve Lawson, post #21670073]

How do I verify my hand calculation quickly?

After solving, check KVL (sum of drops equals the source) and KCL at the node. You can also run a quick simulation; the thread author validated results that way. [Elektroda, Sajed Rakhshani, post #21670080]

What is Ohm’s law and how is it used here?

Ohm’s law states V = I·R. After finding currents from KCL/KVL, multiply by each resistor value to get voltage drops across the resistors. [Elektroda, Cody Gass, post #21670085]