A wire coil is connected to a DC power supply of 20.0 volts. The current in coil

Question

A wire coil is connected to a DC power supply of 20.0 volts. The current in coil is measured to be 775 mA. The same wire coil is then connected to a 180 Hz AC power supply of 20.0 volts (rms). The current in the coil is now measured to be 225 mA (rms). (a) What is the resistance of the wire coil? (b) What is the inductance of the wire coil? (c) Draw a phase diagram showing resistance, reactance, and impedance for the coil when connected to the AC power supply. (d) What is the phase angle between voltage and current in the AC power supply?

Explanation / Answer

a)R = V/I = 20/ 775 x 10^-3 = 25.806 ohms apprx

b) 20 V = Irms (Z)

20 = 225 x 10^-3 (Z)

Z = impedance of circuit= 88.88 ohms

Z^2 = R^2 + w^2 L^2

88.88^2 = 25.806^2 + (2x 3.14 x 180)^2 L^2

7901.234 - 665.949/ 1277804.16 = L^2

L = 75.247 mH

c) Circuit is inductive in nature

d) phase angle = tan^-1 ( wL/ R) = tan^-1 ( (2x 3.14 x 180x  75.247x 10^-3/ 25.806)

phase angle= 73.122 degree

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