Home
Class 12
PHYSICS
A coil of self inductance 10 mH and resi...

A coil of self inductance 10 mH and resistance `0.1Omega` is connected through a switch to a battery of internal resistance `0.9Omega` . After the switch is closed, the time taken for the current to attain 80% of the saturation value is: [take ln 5 = 1.6]

A

`0.016s`

B

`0.324s`

C

`0.002s`

D

`0.103s`

Text Solution

Verified by Experts

The correct Answer is:
A
`i = i_(0) (1 - e^(-Rt//L)) rArr 0.8 i_(0) = i_(0) (1 - e^(-Rt//L))`
`t = (L)/( R) l n5 = (10 xx 10^(-3) xx 1.6)/((0.9 + 0.1)) = 0.016s`
Promotional Banner

Similar Questions

Explore conceptually related problems

A series combination of two resistors 1 Omega each is connected to a 12 V battery of internal resistance 0.4 Omega . The current flowing through it will be

A coil of inductance 8.4 mH and resistance 6 (Omega) is connected to a 12 V battery. The current in the coil is 1.0 A at approximately the time

A coil of inductance 8.4 mH and resistance 6 Omega is connected to a 12 V battery. The current in the coil is 1.0 A at approximately the time.

A coil o finductanc e 8.4mH and resistance 6Omega is connected to a 12V battery. The current in the coil is 1.0A in the time (approx .)

A chock coil has an inductance of 4H and a resistance of 2 Omega . It is connected to a battery of 12 V and negaligible internal resistance. The time taken for the current to become 3.78 A is nearly