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Question: An inductor coil of inductance \[L\] is divided into two equal parts and both parts are connected in...

An inductor coil of inductance LL is divided into two equal parts and both parts are connected in parallel. The net inductance is:
A.LL
B.2L2L
C.L/2L/2
D.L/4L/4

Explanation

Solution

Determine the inductance of the inductors divided into two equal parts. Use the formula for the equivalent inductance of two inductors connected in parallel.

Formula used:
The equivalent inductance Leq{L_{eq}} of two inductors connected in parallel is
1Leq=1L1+1L2\dfrac{1}{{{L_{eq}}}} = \dfrac{1}{{{L_1}}} + \dfrac{1}{{{L_2}}} …… (1)
Here, L1{L_1} is the inductance of the first inductor and L2{L_2} is the inductance of the second inductor.

Complete step by step answer:
The inductor with inductance LL is divided into two equal parts and these two parts are connected in parallel.
Hence, the inductance of both the parts is L2\dfrac{L}{2}.
L1=L2=L2\Rightarrow {L_1} = {L_2} = \dfrac{L}{2}
Determine the equivalent or net inductance of the two inductors connected in parallel.
Substitute L2\dfrac{L}{2} for L1{L_1} and L2{L_2} in equation (1).
1Leq=1L2+1L2\dfrac{1}{{{L_{eq}}}} = \dfrac{1}{{\dfrac{L}{2}}} + \dfrac{1}{{\dfrac{L}{2}}}
1Leq=L2+L2(L2)(L2)\Rightarrow \dfrac{1}{{{L_{eq}}}} = \dfrac{{\dfrac{L}{2} + \dfrac{L}{2}}}{{\left( {\dfrac{L}{2}} \right)\left( {\dfrac{L}{2}} \right)}}
1Leq=LL24\Rightarrow \dfrac{1}{{{L_{eq}}}} = \dfrac{L}{{\dfrac{{{L^2}}}{4}}}
1Leq=4L\Rightarrow \dfrac{1}{{{L_{eq}}}} = \dfrac{4}{L}
1Leq=4L\Rightarrow \dfrac{1}{{{L_{eq}}}} = \dfrac{4}{L}
Leq=L4\therefore {L_{eq}} = \dfrac{L}{4}

Therefore, the net inductance of the two equal parts of inductors connected in parallel is L4\dfrac{L}{4}.

So, the correct answer is “Option D”.

Additional Information:
For the inductors connected in parallel, the net inductance is always less than the smallest inductance of the inductor.

For the inductors connected in series, the voltages though all the inductors are the same.

The net inductance Leq{L_{eq}} for the two inductors connected in series is the sum of the two inductances of the inductors connected in series.
Leq=L1+L2{L_{eq}} = {L_1} + {L_2}

Here, L1{L_1} is the inductance of the first inductor and L2{L_2} is the inductance of the second inductor.

Note:
one may forget to take the reciprocal of the answer obtained by solving the equivalent inductance. If the reciprocal of the answer is not taken, it will be the reciprocal of the net or equivalent inductance and not the net or equivalent inductance.