1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A coil has an inductance of 0.7H and is joined in series with a resistance of 220 Ω. When an alternating e.m.f. of 220V at 50 c.p.s. is applied to it, then the watt less component of the current in the circuit is

A

5 ampere

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

0.5 ampere

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

0.7 ampere

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

7 ampere

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 0.5 ampereThe correct option is (B).Step 1: Given data:Inductance = 0.7HResistance = 220 ΩAlternating e.m.f. = 220V$\mathrm{f}=50\mathrm{cps}$Step 2: Calculation:The impedance of the circuit:Now, $Z=\sqrt{{R}^{2}+{\left(2\mathrm{\pi fL}\right)}^{2}}$$Z=\sqrt{{220}^{2}+{\left(2\mathrm{\pi }×50×0.7\right)}^{2}}$$Z\approx 220\sqrt{2}\mathrm{\pi }$The power factor:$\mathrm{cos}\varphi =\frac{R}{Z}=\frac{\mathrm{\pi }}{4}$The total current in the circuit:$I=\frac{V}{Z}\phantom{\rule{0ex}{0ex}}I=\frac{220}{220\sqrt{2}}\phantom{\rule{0ex}{0ex}}I=\frac{1}{\sqrt{2}}A$Watt less component of current in the circuit:${I}_{a}=I\mathrm{sin}\left(\varphi \right)\phantom{\rule{0ex}{0ex}}{I}_{a}=\frac{1}{\sqrt{2}}×\frac{1}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}{I}_{a}=0.5A$

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Alternating Current Applied to an Inductor
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program