To find the expression for net power consumed, I did this :-

$$ \text{Suppose some resistances } R_1, R_2, R_3, ... \text{ are connected in series in an electric circuit.} \\ \text{Let R be the equivalent resistance. Then} \\ R = R_1 + R_2 + R_3 + ... \\ \text{If 'I' be the current flowing through the circuit, and } V_1, V_2, V_3, ... \text{ be the potential difference across the resistors } R_1, R_2, R_3, ...\text{, then}\\ \frac{V}{I} = \frac{V_1}{I} + \frac{V_2}{I} + \frac{V_3}{I} + ... \\ \text{Multiplying both sides by } I^2, \text{ we get} \\ VI = V_1I + V_2I + V_3 I + ... \\ \implies \boxed{P = P_1 + P_2 + P_3 + ...} $$

But the expression given in the book is

$$ \boxed{\frac{1}{P} = \frac{1}{P_1} + \frac{1}{P_2}+\frac{1}{P_3}+...} $$

What am I doing wrong here ?


Picture from the book :-

enter image description here

  • 1
    $\begingroup$ seems like an error in the book? could you take a picture perhaps? thanks $\endgroup$
    – QuIcKmAtHs
    Jun 12, 2019 at 13:55
  • $\begingroup$ @QuIcKmAtHs I have updated my post (added the picture) $\endgroup$
    – anonymous
    Jun 12, 2019 at 14:22
  • 1
    $\begingroup$ It assumes that the voltage on each resistor is the same as the EMF of the battery. Obviously wrong for a series circuit. Funny book. :) $\endgroup$
    – nasu
    Jun 12, 2019 at 14:42
  • $\begingroup$ What you have done is correct only. In the book $ V^2\over R_1$ is not $P_1$. $P_1$is $ V_1^2\over R_1$ Since the circuit is series. $\endgroup$
    – walber97
    Jun 12, 2019 at 14:42
  • $\begingroup$ What book is this? It seems weird that the conclusion did not look suspect to the author, irrespective of the validity of the calculations. $\endgroup$
    – nasu
    Jun 12, 2019 at 15:41

2 Answers 2


That entire blurb from the textbook doesn't seem too consistent to me.

For starters, lets look at some assumptions the textbook is making. They have a circuit with 3 resistors in series. They show that there is a potential difference $V$ across the whole circuit.

For some reason, they are then saying that each resistor has a voltage $V$ applied across it. This is incorrect. As you have shown, each resistor has it's own voltage drop across the resistor; which are not necessarily equal to each other, and cannot each be equal to $V$ applied to the circuit (see Kirchoff's laws). This means that when they divide both sides of $R = R_1 + R_2 + R_3$ by $V^2$, the $\frac {R_n}{V^2}$ terms don't actually coorespond to $P_n$, because it should be $P_n = \frac {R_n}{V_n ^2}$.

They seem to have gotten mixed up about parallel and series circuits, and instead of saying that $V$ was the same across each resistor, they should have taken $I$ to be the same across each resistor.

If they wanted to use $V$ to determine power, they overcomplicated it, since it should just be $P_{\text{total}} = \frac {V_{\text{total}}^2}{R_{\text{total}}}$

  • $\begingroup$ "They have a circuit with 3 resistors in parallel." No, their diagram show a series circuit. The title of the paragraph as well as the text claim the same thing: series circuit. $\endgroup$
    – nasu
    Jun 12, 2019 at 15:36
  • $\begingroup$ @nasu That's a really common typo for me when talking about series and parallel. Thanks for catching it. $\endgroup$
    – JMac
    Jun 12, 2019 at 15:51

The equation in the book is correct and so is your derivation. The only problem is, your powers (P1,P2...) are not the same. In your derivation, P1 is the actual power output of R1. But in the book, P1 is the rated power, that is the power output by the bulb at voltage V. The conclusion of the book is misleading though.

Say we have bulb 1 rated (P1,V) and bulb 2 rated (P2,V) that is at a voltage V, bulb 1 has a power output of P1 and same for bulb 2. Then, the resistance offered by bulb 1 is V^2/P1 (R1) and by bulb 2 is V^2/P2 (R2) as R = V^2/P. These resistances (R1 and R2) are constant. So when both bulbs are connected in series, the net resistance offered by the bulbs Rnet is R1+R2. Now imagine that we get a new bulb whose resistance is Rnet and whose rated voltage is V. Pnet is the rated power output of this new bulb (that is, Pnet is the power output of the bulb with resistance Rnet at voltage V). So for the new bulb, we can write that Rnet = V^2/Pnet. Since Rnet = R1+R2, we can substitute and say V^2/Pnet = V^2/P1 + V^2/P2. Cancelling the V^2 terms, we get that 1/Pnet = 1/P1 + 1/P2.

The Pnet in the book IS NOT THE NET POWER OUTPUT BY THE 2 BULBS! Rather, Pnet is the power output of a bulb whose resistance is equal to the equivalent resistance of the two given bulbs when connected in series.

Hope it is clearer now.


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