MHT CET · Physics · Electromagnetic Induction
An inductor coil wound uniformly has self-inductance ' \(L\) ' and resistance ' \(\mathrm{R}\) '. The coil is broken into two identical parts. The two parts are then then connected in parallel across a battery of ' \(E\) ' volt of negligible internal resistance. The current through battery at steady state is
- A \(\frac{2 E}{R}\)
- B \(\frac{3 E}{R}\)
- C \(\frac{4 \mathrm{E}}{\mathrm{R}}\)
- D \(\frac{E}{R}\)
Answer & Solution
Correct Answer
(C) \(\frac{4 \mathrm{E}}{\mathrm{R}}\)
Step-by-step Solution
Detailed explanation
In DC, there is no use of inductance.
Since the coil is broken into two identical points, each part will have resistance \(\frac{\mathrm{R}}{2}\).
When these are connected in parallel, their equivalent resistance will be \(\frac{\mathrm{R}}{4}\).
Hence the current \(I\) is given by \(=\frac{E}{R / 4}=\frac{4 E}{R}\)
Since the coil is broken into two identical points, each part will have resistance \(\frac{\mathrm{R}}{2}\).
When these are connected in parallel, their equivalent resistance will be \(\frac{\mathrm{R}}{4}\).
Hence the current \(I\) is given by \(=\frac{E}{R / 4}=\frac{4 E}{R}\)
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