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Question: If the current is halved in a coil, then the energy stored is how much times the previous value...

If the current is halved in a coil, then the energy stored is how much times the previous value

A

12\frac{1}{2}

B

14\frac{1}{4}

C

2

D

4

Answer

14\frac{1}{4}

Explanation

Solution

U=12Li2 6mui.e.U2U1=(i2i1)2=(12)2=14⇒U2=14U1U = \frac{1}{2}Li^{2}\mspace{6mu} i.e.\frac{U_{2}}{U_{1}} = \left( \frac{i_{2}}{i_{1}} \right)^{2} = \left( \frac{1}{2} \right)^{2} = \frac{1}{4} \Rightarrow U_{2} = \frac{1}{4}U_{1}