Question

The S.I. unit of inductance, the Henry can also be written as
$A.$ weber$ \div $ ampere
$B.$ volt second $ \div $ ampere
$C.$ joule $ \div $ ampere$^2$
$D.$ all of the above.

Answer 1

Hint: Here we will proceed by demonstrating all the parts one by one after that we can find out in how many ways S.I. the unit of inductance that is Henry can be written.

Complete step-by-step answer:
The S.I. the unit of inductance is Henry abbreviated as H. It is defined as the measure of electric current changes at one ampere per second resulting in an electromotive force of one volt across the inductor. Henry is a unit based on the 7 base. S.I. unit like meter(m), second(s), kilogram(kg), and ampere (A).
Let us solve part a
We know that self inductance L, is related to flux as
$\phi = L \times I$
$ \Rightarrow L = \dfrac{\phi }{I}$
Where, $\phi = $ weber, and
$I = $ ampere
Therefore, Henry can be written as $L = \dfrac{{weber}}{{ampere}}$
Now for part b, We know that
 $e = L\dfrac{{dI}}{{dt}}$ (where L is constant)
Now, let us consider the units only,
$Volt = Henry\dfrac{{Ampere}}{{Second}}$
It can also be written as
$Volt \times second$ / ampere
Therefore, Henry can also be written as $Volt \times second$ / ampere
Now for part C We know that, the formula for energy dissipation, we will get
$
   = \dfrac{1}{2}l{i^2} \\
   \Rightarrow L = \dfrac{{joule}}{{amper{e^2}}} \\
 $
Therefore, the correct option is D as Henry can be written as weber $ \div $ ampere, $volt - second \div ampere$ , joule$ \div amper{e^2}$


Note: Whenever we come up with this type of problem, one must know that Henry is the S.I. derived unit of electrical inductance. If a current of 1 ampere flowing through the coil produces flux linkage of 1 weber turn, the coil has a self- inductance of 1 henry. By using this approach we can easily solve this question.