an ac generator has a flat coil area A = 40.0cm^2 and N = 100 turns. A motor rot
ID: 2303171 • Letter: A
Question
an ac generator has a flat coil area A = 40.0cm^2 and N = 100 turns. A motor rotates the coil with a constant frequency of 60 rev/s in uniform magnetic field B = 1.50T
a. find the rms voltage Erms generated by coil?
b. the generator is connected to a purely resistive load R = 100 ohms. A dual-channel oscilloscope when connected to circuit, current lags the voltage by 15 degree.
c. to correct the phase difference in part b, would you cinnect a capacitor or an inductor in series with the generator? find the capacitance or inductance that connect to eliminate the phase difference between the voltage and current.
d. assuming the phase difference is corrected, find: the rms current Irms in the circuit, the rms magnetic dipole moment of the coil and then the average torque that msut be applied to the coil to keep it rotating with a constant angular speed.
e. use the result from part d and result P(average) = T(av)*W to find the average power expanded by the motor in rotating coil.
f. find the average power dissipated by the load resistance R (answer should be same with part e)
You must answer all of the parts to receive the points.
Explanation / Answer
w = 60*2*pi/60 = 6.28 rad/s
a) emf_max = N*A*B*w
= 100*40*10^-4*1.5*6.28
= 0.3768 volts
emf_rms = emf_max/sqrt(2)
= 0.3768/sqrt(2)
= 0.2664 volts
b)
c) we need to connect a capacitor
d)
theta = tan^-1(Xc/R)
==> XC = R*tan(theta)
= 100*tan(15)
= 26.8 ohms
impedance of the ckt, z = sqrt(R^2 + Xc^2)
= sqrt(100^2+26.8^2)
= 103.53 ohms
Irms = emf_rms/z
= 0.2664/103.53
= 2.573*10^-3 A
e) Pavg = Irms*emf_rms
= 2.573*10^-3*0.2664
= 6.855*10^-4 Watts
f) Pavg = Irms^2*R
= (2.573*10^-3)^2*100
= 6.62*10^-4 Watts
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