(II) A 75-W incandescent lightbulb is designed to operate with...

Question

(II) A 75-W incandescent lightbulb is designed to operate with an applied ac voltage of 120 V rms. The bulb is placed in series with an inductor L, and this series combination is then connected to a 60.0-Hz 240-V rms voltage source. For the bulb to operate properly, determine the required value for L. Assume the bulb has resistance R and negligible inductance.

Accepted Answer

Welcome back. Everyone in this problem, 100 watt halogen lamp is in series with an inductor L and this combination is powered by a 300 volt RMS 50 Hertz AC voltage source. The lamp is intended to function correctly when an AC voltage of 150 volts RMS is applied across it. Calculate the value of L required for a proper operation assuming the LMS inductance is negligible and it has a resistance R A says it's 0.85 Henry B, 1.2 Henry C 1.5 Henries and D 3.1 Henries. Now, what do we already know here so far? We know that the power of the lamp P is 100 watts. We know that the voltage across the lamp, let's call that VRR MS is 150 volts. OK. We know that the total voltage of the source that is VR MS is 300 volts. And we know that the frequency is 50 Hertz. So how can we use all of this to help us figure out the value of our inductance? L? Well, let's ask ourselves, what do we know about our inductance? Well, we know that our inductance is related to the impedance of the circuit and recall, OK, that the impedance Z is equal to the square root of the resistance squared plus the inductive reactant squared. And our inductive reactants, OK is equal to two pi F multiplied by our inductance. L. So in that case, then that tells us that our impedance is going to be the square root of R squared plus two pi Flr squared. OK. And when we, when we expand that, OK, we can square both sides. That means Z squared is equal to R squared plus two Pi fl all square. If we do a bit of transposition here, that means two pi F squared multiplied by L squared equals Z squared minus R squared. Which means that if we divide both sides by two pi F squared and square root both sides, OK, then L oh OK is gonna be equal to one divided by two pi F multiplied by the square root of Z squared minus R squared. Now, we know how to find L but the problem is so far we don't know OK, what our impedance is. And if we're gonna solve that means we'll need to find our impedance and our resistance. So how can we solve for both of those? Well, let's start by solving for our resistance before we find the resistance, it would be good to find a current. No solving for our resistance came by Ohm's law, we know that the current to the lump is going to be equal to the power divided by the voltage across the lamp. So that's VRR MS. So that is gonna be equal to 100 watts divided by 150 volts, which is gonna be equal to two thirds of an pi and I'm going to leave it this way because I want to get some exact values while I work when I substitute it into the rest of my formula. OK. Now, since we know that this is our current through the lamp, OK. Then again, by S law, that means the resistance is going to be equal to the voltage through the lamp. OK. Divided by the current through the lamp, which is going to be equal to 150 volts divided by two thirds of an empire or multiplied by three halves. OK? Of an umpire which is gonna give us 225 ohms. So our resistance is 225 ohms. Now, what about our impedance? Well, we know that our impedance again, a total impedance is going to be equal to the R MS voltage divided by the current through our circuit, the R MS current. So that's really 300 volts. OK? Divided by two thirds of an ampere. And when we solve, we get it to be equal to 450 ohms. Now that we have the impedance and the resistance we can go ahead and solve for our inductance because by substituting those values, that means our inductance L then is going to be equal to one divided by two pi multiplied by 50 Hertz multiplied by the square root of 450 ohms squared minus 225 ohms square. And now when we go ahead and do that, OK, then we get our inductance L to be equal to 1.24 Henry or to two significant figures, approximately 1.2 Henries. Thus, that tells us that the value of our inductance required for a proper operation is 1.2 Henry. When we look back at our answer, choices, B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

RMS Voltage

Impedance in AC Circuits

Inductive Reactance

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