A uniform wound solenoid coil of self inductance `1.8 xx 10^(-4) H` and resistance ohm is broken into two identical coils. These indentical coils are connected in parallel across a 12 V battery of negligible resistance. Calculate the steady state current through the battery and time constant of the circuit.

Explanation

Here, when the soleniod is broken ino two identical coils, inductance of each of coil becomes half and so is the resistance
`:. L_(1) = L_(2) = (1.8 xx 10^(-4))/(2) = 0.9 xx 10^(-4)` henry
`R_(1) = R_(2) = 6//2 = ohm`.
When these two coils are connected in parallel to a battery, then
`(1)/(L) = (1)/(L_(1)) + (1)/(L_(2)) = (L_(2) + L_(1))/(L_(1) L_(2))`
`= 0.45 xx 10^(-4) H`
and `R = (R_(1) R_(2))/(R_(1) + R_(2)) = (3 xx 3)/(3 + 3) = 1.5 phm`
Time constant,
`tau = (L)/(R ) = (0.45 xx 10^(-4))/(1.5) = 3 xx 10^(-5) s`
The steady state current does not depend upon L.
`I_(0) = (E)/(R ) = (12)/(1.5) = 8 A`