wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Two indicators L1 (inductance 1 mH, internal resistance 6 Ω) and L2 (inductance 2 mH, internal resistance 6 Ω), and a resistor R (resistance 6 Ω) are all connected parallel across a 12 V battery. The circuit is switched on at time t=0. The ratio of the maximum to minimum current (Imax/Imin) drawn from the battery is (only integer)

Open in App
Solution


At t=0, inductance acts as a resistor of infinite resistance.

Hence, Reff=R1=R=6 Ω.

At steady state, inductors offer zero resistance, hence,

1Reff=1R2=16+16+16

R2=2 Ω

Hence, the current will be maximum at steady state and minimum at t=0.

So, Imax=ER2=122=6 A and

Imin=ER1=126=2 A

ImaxImin=62=3

Correct answer:3
Why this question ?
Key trick: At the time when the current starts flowing, we will have a minimum current at t=0. In this time the whole current will pass through resistor only and no current would be passing through inductors. After a long time, current will also pass through inductors, and we will have our maximum current.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Equivalent Resistance in Series Circuit
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon