First slide

Phasors and Phasor Diagrams

Question

Two resistors are connected in series across 5 volt r.m.s. source of alternating potential as shown in figure. The potential difference across $6 Ω$ resistor is 3V. If R is replaced by pure inductor L of such magnitude that current remains same, potential difference across L is

Moderate

A

1 V
B

2 V
C

3 V
D

4 V

Solution

$V_{rms} (6 Ω) = 3 V = 6 I_{V} ∴ I_{V} = 0 .5 A$
$I_{V} = \frac{1}{2} = \frac{5}{\sqrt{6^{2} + X_{L}^{2}}} \Rightarrow X_{L} = 8 Ω$
Now, $V_{L} = I_{V} \cdot X_{L} = \frac{1}{2} \times 8 = 4 V$

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