A uniform wound solenoidal coil of self inductance `1.8xx10^(-4)` henry and resistance 6 ohm is broken up into two identical coils. These identical coils are then connected in parallel across a 15-volt battery of negligible resistance. The time constant for the current in the circuit is.......seconds and the steady state current through the battery is ..............amperes.

Explanation

The coil is broken into two indentical coils.
`L_(eq) = (L//2 xx L//2)/(L//2 + L//2) = (L)/(4) = 0.45 xx 10^(-4) H`,
`R_(eq) = (R//2xx R//2)/((R )/(2) + (R )/(2)) = (R )/(4) = 1.5 Omega`
Time constant `= (L_(eq))/(R_(eq)) = (0.45 xx 10^(-4))/(1.5) = 0.3 xx 10^(-4) s`. Steady
Current,
`I = (E)/(R_(eq)) = (12)/(1.5) = 8A`