A coil of inductance 0.1 H is connected to an alternating voltage generator of voltage E = 100 sin (100 t) volt. The current flowing through the coil will be

I = 10 \sqrt{2} s i n (100 t) A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I = 10 \sqrt{2} c o s (100 t) A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I = - 10 s i n (100 t) A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I = - 10 c o s (100 t) A

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B $I = - 10 c o s (100 t) A$
L = $0.14$
E= $100 s i n (100 t)$
$\Rightarrow X_{L} = w_{L} = 100 \times 0.1$
$= 10 Ω$
$\Rightarrow i_{\circ} = \frac{E \circ}{X_{L}} = \frac{100}{10} = 10 A$
$\Rightarrow i = i_{\circ} sin (100 t - \frac{π}{2})$
$\Rightarrow i = - i_{\circ} sin (\frac{π}{2} - 100)$
So we get
$\Rightarrow i = - 10 cos (100 t) A$
therefore, option $D$ is correct.

Suggest Corrections

Similar questions

5.0 m H

V = 10 sin (100 t)

E = 100 sin 100 π t, I = 5 sin 100 π t

I = 10 sin (100 π t)

5 π

V (t) = 100 s i n (500 t)

L = 0.02 H .

10 Ω

100 m H

V = 100 cos (100 t)

Join BYJU'S Learning Program