# Equivalent power of a series and parallel combination of resistors [closed]

How do you derive the equivalent power for series combination and parallel combination of resistors? I seem to get the same expression for both. This is what I've done:

Series

Current $I$ is a constant in series combination

$\hspace{1.3cm} P_1=I^2R_1$

$\hspace{1.3cm}P_2=I^2R_2$

$\hspace{1.3cm} P_1+P_2=I^2(R_1+R_2)\\ \implies P_{eff}=I^2R_{eff}$

Parallel

Voltage $V$ across the resistors are taken to be a constant

$\hspace{1.3cm} P_1=\frac{V^2}{R_1}$

$\hspace{1.3cm}P_2=\frac{V^2}{R_2}$

$\hspace{1.3cm} P_1+P_2=V^2(\frac{1}{R_1}+\frac{1}{R_2})\\ \implies P_{eff}=\frac{V^2}{R_{eff}}$

In both cases I get the same expression for $P_{eff}$. Obviously this is wrong. But I'm not sure where I'm going wrong. Help is appreciated.

## closed as off-topic by rob♦Jun 15 '18 at 15:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – rob
If this question can be reworded to fit the rules in the help center, please edit the question.

• How do you know it's wrong? – David Z Aug 2 '16 at 16:53
• Because, $P_{eff}=\frac{V^2}{R_{eff}}=I^2R_{eff}, i.e power in parallel combination and series combination are the same. Shouldn't parallel combination be less than series combination? – abhijeetviswa Aug 2 '16 at 17:20 ## 1 Answer Your derivation is right. why do you think it is wrong? Because you can manipulate to make the second expression same as the first one and since one is in series and the other is parallel, so they must be wrong? But you arrived at the two expressions through different paths, in the first you took current as constant( series combination) and in the second you took pot. diff. as constant ( parallel combination) . In any combination of resistors, after you find out the equivalent resistance , the total current across the combination and the total potential drop across the combination, you can write the effective power as any of the above two expressions ( given by you). But if you start from individual resistors, you will have to follow diff. paths for different types of combinations, as you have followed. But if you have calculated the equivalent resistance , it does not matter( except while calculating the eq. resistance) how they are combined, the combination can be essentially considered as a single resistor and so the effective power will essentially have the same set of interchangable expression for any single resistor ( originating from a combination of resistors or not). Hope it helps • Well, I did this for my exam and I was awarded no marks. So, is my equation correct because$R_{eff}$in series and parallel will give me different values? – abhijeetviswa Aug 2 '16 at 17:17 • yes, i think it's correct. I can't say about your exam as i don't know the exact question in your paper and what answer the teacher expected ( since some teachers are biased towards their notes XD ). – D.K. Aug 2 '16 at 17:24 • The question is as follows: Two heating elements of resistance R1 and R2 when operated at a constant supply of voltage V, consume power P1 and P2 respectively. Deduce an expression for the power of these combination when they are in turn connected(1) in series and (2) in parallel across the same voltage supply. And, the notes given by my teacher say that$\frac{1}{P_{series}}=\frac{1}{P_1}+\frac{1}{P_2}$and$P_{parallel}=P_1+P_2\$ – abhijeetviswa Aug 2 '16 at 17:33
• Your answer seems right, same as your teacher's notes. Maybe your answer was not clear enough on paper or you did not use the terms P1 and P2 in the end expression, i don't know. Conceptually u are correct. Talk to your teacher about it, maybe fetch yourself some marks. cheers – D.K. Aug 2 '16 at 17:38