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

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    $\begingroup$ seems like an error in the book? could you take a picture perhaps? thanks $\endgroup$ – QuIcKmAtHs Jun 12 at 13:55
  • $\begingroup$ @QuIcKmAtHs I have updated my post (added the picture) $\endgroup$ – arandomguy Jun 12 at 14:22
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    $\begingroup$ 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 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 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 at 15:41

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 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 at 15:51

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