# Expression for net power consumed when resistances are connected in series in electric circuit

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 :-

• seems like an error in the book? could you take a picture perhaps? thanks – QuIcKmAtHs Jun 12 at 13:55
• @QuIcKmAtHs I have updated my post (added the picture) – arandomguy Jun 12 at 14:22
• 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. :) – nasu Jun 12 at 14:42
• 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. – walber97 Jun 12 at 14:42
• What book is this? It seems weird that the conclusion did not look suspect to the author, irrespective of the validity of the calculations. – nasu Jun 12 at 15:41

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