First slide

Phasors and Phasor Diagrams

Question

An alternating voltage is connected in series with a resistance R and an inductance L. If the potential drop across the resistance is 200 V and across the inductance is 150 V, then the applied voltage is

Easy

A

350 V
B

250 V
C

500 V
D

300 V

Solution

The applied voltage is given by $V = \sqrt{V_{R}^{2} + V_{L}^{2}}$
$V = \sqrt{(200)^{2} + (150)^{2}} = 250 volt$

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