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The value of resistance is 3Ω and inductor impedance is j4Ω.

As far as I have learned the wattmeter measures instantaneous power but shows reading for average power. As I solved for the given values the power comes out 12W in both cases even though they are connected at different points in the circuit and I want to know why?

I can't seem to figure out the reason behind this. Can someone please point out? Thank youImage for reference of circuits in both cases

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  • $\begingroup$ Hi Darkco, and welcome to Physics Stack Exchange! Could you expand on this a bit? Why does it surprise you that the results are the same? Did you expect them to be different? (If so, why?) Could you also include a summary of the calculations you did to find out that they are the same? $\endgroup$ – David Z May 24 at 5:48
  • $\begingroup$ @DavidZ As far as I have learned the wattmeter measures average power. In case 1 it is connected across a resistor and in case 2 it is connected across the combination of resistor and inductor. So I thought the power readings will vary, however they are not. $\endgroup$ – Darkco May 24 at 6:20
  • $\begingroup$ I meant edit it into your question. It's not much help to leave information like this in a comment, which may disappear later, but if you put it in the question, it will be available to all future readers. (Note that we don't like to have it added in at the end labeled with "EDIT:" or anything like that. Change the question to include this information so it looks like it had always been written that way. If people want to know what changed, we do have a revision history available.) $\endgroup$ – David Z May 24 at 7:06
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(a) Wouldn't you expect this result? A pure inductor takes in no mean power over a period of time, so the power taken in by the inductor-resistor combination ought to be the same as that dissipated in the resistor alone.

Looking at this in terms of current and voltage, let the circuit current be $$I(t)=I_0 \cos (\omega t)$$

Then the pd across the resistor is $$V_R(t)=I_0 R \cos (\omega t)$$ and that across the inductor is $$V_L(t)=-I_0 \omega L \sin (\omega t)$$ So for your first connection of the wattmeter the instantaneous power is $$I(t) V(t) = I(t) V_R(t)= I_0^2 R \cos^2 (\omega t)$$

And for your second connection of the wattmeter the instantaneous power is $$I(t) V(t) = I(t) [V_R(t) + V_L(t)] = I_0^2 R \cos^2 (\omega t)– I_0^2 \omega L \cos (\omega t) \sin (\omega t)$$ The second term averages to zero over any whole number of cycles, leaving you with just the power in the resistor.

(b) How does the wattmeter work? Essentially the circuit current, $I(t)$ is carried in a low impedance path through the meter between the top two terminals, while the voltage, $V(t)$ is registered between the left hand terminal and the bottom right terminal (between which the meter impedance is high). The meter calculates $I(t)V(t)$ (instantaneous power) at frequent intervals and presents you with its mean value over a period of time. This is the mean power.

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  • $\begingroup$ Thanks a lot for clearing my doubt. Good day sir! $\endgroup$ – Darkco May 26 at 17:13

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