Answer to Question #126678 in Physics for Susan Williams

Question #126678
An inductor, L, of 2 H and a resistor, R, of 1000Ω are in series across an a.c. voltage of 10 V, frequency 60 Hz. Calculate the current in r.m.s. flowing,
1
Expert's answer
2020-07-20T14:56:32-0400

According to the Ohm's law, the current amplitude is:


I=UZI = \dfrac{U}{Z}

where U=10VU = 10V is the volage amplitude and ZZ is the total impedance of the circuit.

The impedance of the series resistor and inductor is:


Z=R2+(2πfL)2Z = \sqrt{R^2 + (2\pi fL)^2}

where f=60Hzf = 60Hz is the frequency of a.c. voltage.

Substituting the numerical values, get:


I=UR2+(2πfL)2=1010002+(2π602)20.00798AI = \dfrac{U}{\sqrt{R^2 + (2\pi fL)^2}} = \dfrac{10}{\sqrt{1000^2 + (2\pi\cdot 60\cdot 2)^2}} \approx 0.00798A

The r.m.s. of a sinusoidal current is:


Irms=I20.00564AI_{rms} = \dfrac{I}{\sqrt{2}}\approx 0.00564A

Answer. 0.00564 A = 5.64 mA.


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