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A large electromagnetic coil is connected to a 120-Hz ac source. The coil has resistance 400 Ω, and at this source frequency the coil has inductive reactance 250 Ω. (a) What is the inductance of the coil? (b) What must the rms voltage of the source be if the coil is to consume an average electrical power of 450 W?

Short Answer

Expert verified

a) Inductance of the coil is 0.33 H

b) Rms voltage of the source is 500.31 V

Step by step solution

01

Concept of reactance, impedance and average power

Inductive reactance is the resistance offered by an inductor in a circuit. It is denoted asXL and is calculated using the formula, XL=ωL, where,ω is the angular frequency of the A/C signal, andL is the inductance.

Impedance of a circuit is the total resistance offered by the circuit.Denoted by Z, the formula for impedance of a LR circuit is ZR2+XL2.

Average power of an A/C circuit is given by the formula Pavg=VrmsIrscosϕ, wherePavg is the average power,Vrms is the rms value of the voltage,Irms is the rms value of current,ϕ is the phase difference between voltage and current. It is known that cosϕ=RZ, and Irms=VrmsZ. Using these two,Pavg the formula for comes out to be-Pavg=VrmsRZ2 .

02

Calculation of inductance of the coil

Given that-

XL=250Ωω=2π×120=753.98rad/sVrms=PavgZ=450×471.7=145.69V

Now, from previous discussion it is known that, XL=ωL. Therefore,

L=XLω=250753.98=0.33H

Hence, the inductance of the coil is 0.33 H.

03

Calculation of rms value of voltage

The impedance of the circuit is can be given by,Z=R2+XL2, therefore,

Z=4002+2502=471.7Ω

Now, average power is given by, Pavg=Vrms2RZ2, therefore,

Vrms=ZPavgR=471.7×450400=500.31V

Therefore, rms value of voltage is 500.31 V

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