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

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Correct answer: A. 30 μ s, 8 ~A

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 μ 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

A uniform coil of self-inductance 1.8 × 10^ -4 H and resistance 6...

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