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I knew that power of two resistors in series is equal to $$\frac{P_1P_2}{P_1+P_2}$$ where $P_1$ and $P_2$ are the individual powers of the two resistors. I tried to prove it using $P=I^{2}R$ and I took out the value of $R_1$ and $R_2$ using that equation. I substituted the values of $R_1$ and $R_2$ in $R_3=R_1+R_2$ where $R_3$ is the net resistance in series circuit. The net power is $P_3=I^2R_3$, $R_1=P_1/I^2$, $R_2=P_2/I^2$, and $R_3=P_3/I^2$. On substituting it in $$R_3=R_1+R_2$$ we get $$P_3=P_1+P_2 \, .$$ Where am I wrong? Or is the formula $$P_3=\frac{P_1P_2}{P_1+P_2}$$ applicable only when both the resistors have same voltage? Can you tell me how to solve this?

We now that $P=\frac{P_1P_2}{P_1+P_2}$ if we use $P=\frac{V^2}{R}$. But we also know that $P=I^2R$, so we get $P=P_1+P_2$ (proved in the question ) . So can we equate the two powers to get $$\frac{P_1P_2}{P_1+P_2}=P_1+P_2$$ On simplifying it , we get $$P_1^2+P_2^2+P_1P_2=0$$ This implies that $$P_1=P_2(-1+i\sqrt 3)$$ and $$P_1=P_2(-1-i\sqrt 3)$$ But this is not always true .

Also if we multiply $P=\frac{P_1P_2}{P_1+P_2}$ and $P=P_1+P_2$, we get $$P^2=P_1P_2$$ This means $$P=+\sqrt{P_1P_2}$$

Is this correct ?

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  • $\begingroup$ You can use $<mathjax>$ to get inline math, instead of using $$<mathjax>$$ which puts the equation on a separate line. $\endgroup$ – The Photon Jun 1 '18 at 15:51
  • $\begingroup$ Re your recently added section: you cannot write $\frac{P_1P_2}{P_1 + P_2}= P_1+P_2$ because the $P_i$ on the left side have a different meaning than the $P_i$ on the right side. On the left side, $P_i = \frac{V^2_S}{R_i}$ while on the right side, $P_i = I^2R_i = \left(\frac{V_S}{R_1 + R_2}\right)^2R_i$. You should instead write something like $$ \frac{P_1P_2}{P_1 + P_2}= P'_1+P'_2$$ $\endgroup$ – Alfred Centauri Jun 5 '18 at 12:18
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I'm not sure where your first equation comes from. The total power delivered to two resistors, whether series connected or parallel connected is just the sum of the individual resistor powers.

Series connected resistors have identical current through and so the total power delivered to two series connected resistors is just the power delivered to the equivalent resistance:

$$P = P_1 + P_2 = I^2_1R_1 + I^2_2R_2 = I^2(R_1 + R_2) = I^2R_{eq},\quad R_{eq} = R_1 + R_2$$

Parallel connected resistors have identical voltage across and so the total power delivered to two parallel connected resistors is just the power delivered to the equivalent resistance:

$$P = P_1 + P_2 = \frac{V^2_1}{R_1} + \frac{V^2_2}{R_2} = V^2\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{V^2}{R_{eq}},\quad \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$$


This is given in some books and was asked in NTSE examination

Perhaps you've misunderstood the question that the given formula is the answer to?

Let's try a different question instead. Here's the setup:

  • $P_1$ is the power delivered to $R_1$ when connected across a voltage $V$
  • $P_2$ is the power delivered to $R_2$ when connected across a voltage $V$

The question is then: what is the power $P$ delivered to the series combination of $R_1$ and $R_2$ when connected across a voltage $V$?

The answer is:

$$P = \frac{V^2}{R_1 + R_2}$$

But

$$P_1 = \frac{V^2}{R_1}$$

$$P_2 = \frac{V^2}{R_2}$$

$$\frac{P_1P_2}{P_1 + P_2} = \frac{\frac{V^4}{R_1R_2}}{V^2(\frac{1}{R_1} + \frac{1}{R_2})} = \frac{V^2}{R_1 + R_2} = P$$

To summarize, the answer depends on how what one means by "individual powers of the two resistors".

If $P_1$ and $P_2$ are the individual powers of the series connected resistors, then $P = P_1 + P_2$ as I wrote in the first part of my answer.

However, if $P_1$ and $P_2$ are the respective resistor powers when connected alone across a voltage source, the power delivered to the series combination is $P = \frac{P_1P_2}{P_1 + P_2}$

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  • $\begingroup$ This is given in some books and was asked in NTSE examination $\endgroup$ – MathsWiz Jun 2 '18 at 2:12
  • $\begingroup$ @MathsWiz, I've updated my answer. $\endgroup$ – Alfred Centauri Jun 2 '18 at 13:32
  • $\begingroup$ @MathsWiz Could you please report the exact wording of those books (at least one of them) and specify their titles? $\endgroup$ – Massimo Ortolano Jun 3 '18 at 19:21
  • $\begingroup$ It is an exam preparation book with the title NTSE Supplement 1. But you can't buy it because it is specially prepared by some coaching institutes. The exact wording of that book is "Two resistors of powers $P_1 $ and $P_2$ are connected in series. Calculate the net power supplied by them ." $\endgroup$ – MathsWiz Jun 4 '18 at 4:19
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    $\begingroup$ @MathsWiz, thank you for the additional context. Are you certain about the wording? In particular, I'm concerned with "Calculate the net power supplied by them." (Resistors don't supply power - resistors convert electrical energy to heat) $\endgroup$ – Alfred Centauri Jun 5 '18 at 1:16

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