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# When induced emf in inductor coil is 50% of its maximum value then stored energy in inductor coil in the given circuit will be-A. 2.5 mJB. 5 mJC. 15 mJD. 20 J

Last updated date: 07th Aug 2024
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Hint: Inductor is a device which responds to the change in the current. It stores the magnetic energy. When an inductor is connected to the power source initially it won’t allow the flow of current but after some time it will act as a normal conducting wire with no resistance which will be exactly opposite in case of capacitor.

Formula used:
$E = \dfrac{1}{2}L{i^2}$
$V = iR$

Normally in case of capacitors initially when power source is connected to the capacitor it acts as
a normal conduction wire and after sometime this will completely not allow the flow of charge but in case of an inductor this will be completely opposite. When an inductor is connected to the power source initially it won’t allow the flow of current but gradually after some time it will allow the flow of current just as the normal conducting wire.
So if we observe the diagram given the maximum voltage across the inductor will be at the starting. Hence the maximum voltage will be nothing but the emf of the battery as initially the battery and inductor will be parallel. So the maximum voltage is 2 volts.
50 percent of this voltage will be 1 volt.

Sum of voltages across the inductor and one ohm resistor must be 2 volts. So voltage across 1 ohm resistor will be 2-1 i.e 1 volt
From ohm’s law the current in the circuit will be
$V = iR$
\eqalign{ & \Rightarrow 1 = i1 \cr & \Rightarrow i = 1A \cr}
So energy stored in the inductor will be
$E = \dfrac{1}{2}L{i^2}$
\eqalign{ & \Rightarrow E = \dfrac{1}{2}5{(1)^2} \cr & \Rightarrow E = \dfrac{5}{2} \cr & \Rightarrow E = 2.5mJ \cr}

So, the correct answer is “Option A”.

Note:
In case of an inductor when the circuit is closed then there is sudden increase in the flow of current which means that sudden change will be opposed by the inductor. Hence inductor generates emf itself in such a way that it opposes that flow and hence net current flown will be zero and initially inductor doesn’t allow the current.