Question

What inductance would be needed to store\[1.0Kwh\] of energy in a coil carrying current a \[200A\] current?
\[\begin{align}
  & A)90H \\
 & B)90mH \\
 & C)180H \\
 & D)180mH \\
\end{align}\]

Answer 1

Hint: In this numerical we have to find the value of inductance and for that purpose we have to use the relation to inductance and energy and using that result we get the desired value of inductance. In this relation energy should be used in joule.SI unit of inductance is Henry that should be used with its magnitude.

Complete step by step solution:
In Induction, current is produced either on the same coil or on the neighboring coli when a changing current is passed through it. When induced current is produced in the same coil on which changing main current is passing the phenomenon of self induction is done and when induced current is produced on the other coil then it is called mutual induction. The direction of induced current is found by the right hand thumb rule. Different coefficients of induction are used in self and mutual induction like \[L\] is used in self induction phenomenon and \[M\] is used in mutual induction phenomenon. Both the phenomenon occurs on the same coil also because as soon as changing current starts passing self induction is produced and if any neighboring coli is placed then mutual induction also starts. Mutual inductance varies depending upon the orientation and positions of two coils.
In the given question we have to find the value of self induction and values of energy and current are given.
Since energy and inductance is connected by a relation –
\[U=\dfrac{1}{2}L{{i}^{2}}\]
Where,
\[\begin{align}
  & U=Energy \\
 & L=Induc\tan ce \\
 & i=current \\
\end{align}\]
Since in the given question energy is given as \[1.0Kwh\].
First it will convert into joules by using the below conversion
\[1Kwh=3.6\times {{10}^{6}}J\].
So, energy becomes in joules as,
\[E=3.6\times {{10}^{6}}J\]
\[\begin{align}
  & i=200A \\
 & L=? \\
\end{align}\]
Now we put these value in above formula we get,
\[3.6\times {{10}^{6}}=\dfrac{1}{2}\times L\times {{(200)}^{2}}\]
\[\Rightarrow 7.2\times {{10}^{6}}=L\times 40000\]
\[\begin{align}
  & \Rightarrow L=\dfrac{7200000}{40000} \\
 & \therefore L=180H \\
\end{align}\]
The value of inductance needed to store \[1.0Kwh\] of energy in a coil carrying current a \[200A\] current is \[180H\]. The SI unit of inductance is Henry and it is represented by the symbol \[H\].
So, the correct option is C.

Note: In both the phenomena of self induction and mutual induction, magnetic flux is directly proportional to the current and as current will change; magnetic flux will also change and due to this change in magnetic flux an induced e.m.f. is set up and it produces induced current whose direction is identified by Lenz law.