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An inductor of 0.5 H is connected in series with a resistance of 1000. The voltage supplied to the junction is expressed by v = 400 sin (2π 50 t + 45°) then - (i) Write the equation of the circuit current and (ii) Find the power factor.

An inductor of 0.5 H is connected in series with a resistance of 1000. The voltage supplied to the junction is expressed by v = 400 sin (2π 50 t + 45°) then - (i) Write the equation of the circuit current and (ii) Find the power factor.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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An inductor of 0.5 H is connected in series with a resistance of 1000. The voltage supplied to the junction is expressed by v = 400 sin (2π 50 t +45°). then - (i) Write the equation of the circuit current and (ii) Find the power factor. 45° 12.5° I
Transcribed Image Text:An inductor of 0.5 H is connected in series with a resistance of 1000. The voltage supplied to the junction is expressed by v = 400 sin (2π 50 t +45°). then - (i) Write the equation of the circuit current and (ii) Find the power factor. 45° 12.5° I
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