Question

The equivalent inductance of two inductors is $2.4H$ when connected in parallel and \[10H\] when connected in series. What is the value of inductance of the individual inductors?
\[\begin{align}
  & A.8H,2H \\
 & B.6H,4H \\
 & C.5H,5H \\
 & D.7H,3H \\
\end{align}\]

Answer 1

Hint: The formula for series and parallel condition of inductance is similar to that of resistance. In series the inductance is the sum of inductance and in parallel resultant inductance is the sum of the reciprocal of individual inductance.

Complete step by step answer:
To proceed further, let us know about Inductance and its types.
Inductance: We know that the magnetic flux in any conductor is directly proportional to the current.
\[{{\phi }_{B}}\propto I\]
Where, \[{{\phi }_{B}}\] is the magnetic flux and \[I\]is the current.
\[\begin{align}
  & N{{\phi }_{B}}\propto I \\
 & N{{\phi }_{B}}=LI \\
\end{align}\]
Where, \[N\] is the number of turns in the induction coil. And here \[L\] is the constant which is the ratio of net flux to the electric current and is known as Inductance.
Inductance is a scalar quantity and its SI unit is Henry or H.
In series: When two coils of inductance \[{{L}_{1}}\] and \[{{L}_{2}}\]connected in series. Then inductance will add:
\[{{L}_{series}}={{L}_{1}}+{{L}_{2}}\]
In Parallel: When two coils of inductance \[{{L}_{1}}\] and \[{{L}_{2}}\]connected in parallel. Then reciprocal of inductance will add:
\[\dfrac{1}{{{L}_{parallel}}}=\dfrac{1}{{{L}_{1}}}+\dfrac{1}{{{L}_{2}}}\]
In our question it is given that:
\[\begin{align}
  & {{L}_{parallel}}=2.4H \\
 & {{L}_{series}}=10H \\
\end{align}\]
Let \[{{L}_{1}}\]and \[{{L}_{2}}\]be the two individual inductor, therefore:
\[\begin{align}
  & \dfrac{1}{{{L}_{1}}}+\dfrac{1}{{{L}_{2}}}=\dfrac{1}{2.4} \\
 & \dfrac{1}{{{L}_{1}}}+\dfrac{1}{{{L}_{2}}}=\dfrac{10}{24} \\
 & \dfrac{{{L}_{1}}+{{L}_{2}}}{{{L}_{1}}{{L}_{2}}}=\dfrac{10}{24} \\
\end{align}\]
Also,
\[{{L}_{1}}+{{L}_{2}}=10\]
Or
\[{{L}_{2}}=10-{{L}_{1}}\]
And,
\[\begin{align}
  & \dfrac{10}{{{L}_{1}}{{L}_{2}}}=\dfrac{10}{24} \\
 & {{L}_{1}}{{L}_{2}}=24 \\
\end{align}\]’
\[\begin{align}
  & {{L}_{1}}(10-{{L}_{1}})=24 \\
 & {{L}^{2}}_{1}-10{{L}_{1}}+24=0 \\
 & ({{L}_{1}}-6)({{L}_{1}}-4)=0 \\
 & {{L}_{1}}=6,{{L}_{1}}=4 \\
\end{align}\]
Or
\[{{L}_{2}}=6,{{L}_{2}}=4\]
Hence the inductance of two individual inductors are \[6H\]and \[4H\] respectively.
Therefore the correct answer is option B.

Additional Information:
Types of Inductance: There are two types of Inductance.
Self Inductance: When current passing through the coil changes, magnetic flux also changes hence induced EMF is produced in the same coil. This phenomenon is called self inductance.
Mutual Inductance: When the current changes in a coil, then flux linked with the other coil placed near the first changes. Hence an induced current is produced in the second coil. This phenomenon is called mutual inductance.
The first coil in which current changes is called primary coil and the coil in which flux changes is called secondary coil.
The function of a transformer is based on the principle of mutual induction.

Note: If the question is objective type as of above. Then students can directly put the values of options to the condition of parallel and series. You will get the answer quickly. Let us see how:
The sum of every value in options gives us \[10H\]which will not give us the right answer. Now let's put the condition in parallel.
\[{{L}_{parallel}}=\dfrac{{{L}_{1}}{{L}_{2}}}{{{L}_{1}}+{{L}_{2}}}\]
Option B, is satisfying the above relation. Hence option B is correct.