First slide

Inductance

Question

A coil resistance $20 Ω$ and inductance 5 H is connected with a 100 V battery. Energy stored in the coil will be

Easy

A

41.5 J
B

62.50 J
C

125 J
D

250 J

Solution

$U = \frac{1}{2} {Li}^{2} = \frac{1}{2} L {(\frac{E}{R})}^{2} = \frac{1}{2} \times 5 \times {(\frac{100}{20})}^{2} = 62.50 J$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App