1
$\begingroup$

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 ?

Edit:

Picture from the book :-

enter image description here

$\endgroup$
  • 1
    $\begingroup$ 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
  • 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 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
1
$\begingroup$

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 $P_{\text{total}} = \frac {V_{\text{total}}^2}{R_{\text{total}}}$

$\endgroup$
  • $\begingroup$ "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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.