Question

A coil of ${10}^{-2}$ H inductance carries a current I = 2sin (100t)A. When current is half of its peak value then at that instant the induced emf in coil will be

• A. 1 V
• B. $\sqrt{2}V$
• C. $\sqrt{3}V$
• D. 2 V

Answer 1

The correct option is C $\sqrt{3}V$

Given that,

$I=2sin\left(100t\right)$… (I)

Now,

$i_{max}=2A$

$\frac{i_{max}}{2}=1A$

We know that,

$1=2\sin \left(100t\right)$

$\sin \left(100t\right)=\frac{1}{2}$

$100t=\frac{\pi }{6}$

Now,

Differentiate of equation (I)

$I=2\sin \left(100t\right)$

$\frac{dI}{dt}=2\times 100\cos \left(100t\right)$

$\frac{dI}{dt}=200\times \cos \left(\frac{\pi }{6}\right)$

$\frac{dI}{dt}=200\times \frac{\sqrt{3}}{2}$

$\frac{dI}{dt}=100\sqrt{3}$

Now, the induced e m f in coil

$e_{ind}=L\times \frac{dI}{dt}$

$e_{ind}={10}^{-2}\times 100\sqrt{3}$

$e_{ind}=\sqrt{3}$

Hence, the induced e m f in coil is √3 volt