A uniform coil of self-inductance 1.8 10⁻⁴ H and resistance 6 is broken up into two identical coils. These two coils are then connected in parallel across a 12 ~V battery. The circuit time constant and steady state current through the battery respectively are

Step-by-step solution

When coil is broken into two identical parts, then resistance of each part R^ = R 2 = 6 2 =3 Inductance of each part L^ = L 2 = 1.8 2 10⁻⁴=0.9 10⁻⁴ H Now, time constant = L^ R^ = L 2 R 2 = L R = 1.8 10⁻⁴ 6 =30 s Now, effective resistance when both coils are connected in parallel R^ = R^ R^ R^ +R^ = 3 3 3+3 = 9 6 So, maximum current drawn from battery I= V R^ = 12 9 / 6 =8 ~A