Question 429: AC fundamentals

Question

An inductor coil wound uniformly has self inductance 
 'L' and resistance 'R'. The coil is broken into two identical parts. The two parts are then connected in parallel across a battery of 'E' volt of negligible internal resistance. The current through battery at steady state is

Accepted Answer

Since the coils are connected to a dc voltage. Inductive reactance will be zero, only resistance has to be considered.
When the coil is broken into two identical points, each part will have resistance $$\mathrm{\frac{R}{2}}$$. When these are connected in parallel, their equivalent resistance will be $$\mathrm{\frac{R}{4}}$$. Hence the current I is given by $$\mathrm{I=\frac{E}{R/4}=\frac{4E}{R}}$$

Answer

$$\mathrm{\frac{2E}{R}}$$

Answer

$$\mathrm{\frac{3E}{R}}$$

Answer

$$\mathrm{\frac{4E}{R}}$$

Answer

$$\mathrm{\frac{E}{R}}$$

AC fundamentals

An inductor coil wound uniformly has self inductance 'L' and resistance 'R'. The coil is broken into two identical parts. The two parts are then connected in parallel across a battery of 'E' volt of negligible internal resistance. The current through battery at steady state is

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