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$\bullet($ a) Show that for an $R-L-C$ series circuit the power factor is equal to $R / Z$ . (Hint: Use the phasor diagram; see Figure 22.13 $\mathrm{b} .$ (b) Show that for any ac circuit, not just one containing pure resistance only, the average power delivered by the voltage source is given by $P_{\mathrm{av}}=I_{\mathrm{mm}}^{2} R .$

a) $\frac{R}{Z}$b) $P=I_{r m s}^{2} R$

Physics 102 Electricity and Magnetism

Chapter 22

Alternating Current

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Electromagnetic Induction

Rutgers, The State University of New Jersey

University of Washington

Hope College

Lectures

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$\bullet$ In a series $R-L…

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you knew this problem, we're going to be looking at the face or diagram that they mentioned in the problem. And so from that diagram, I see the co sign of the phase angle fly. So this is the power factor is equal to the voltage across the resistor over the total voltage. And so now I can use owns a law to relate these two voltages to the currents and that can cancel occurrence and keep the resistance is and the Indians and so v r is equal to current times resistance. And then the total B is equal to current times and peons. Then we can cancel the currents when we get that, which is what we want to show in the first place. And probably we're going to start with this formula. Yeah, which power is equal to the Armas voltage times? The army's current times, the power factor. We know what the power factor is, so let's go ahead and substitute that and with them. And then now I'm going to move this impedance over into this term. And so if I just erase this here, that impedance is going to go with the voltage our mess like that. And then now I'm going to use the fact that the voltage armistice over the impudence is equal to the current armistice. And you'll see, whenever I multiply this by this, it's gonna square it. I'll get I squared R and this is what we want show and so

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