The equation of current in a purely inductive circuit is $5 \sin \left(49 \pi \mathrm{t}-30^{\circ}\right)$. If the inductance is 30 mH then the equation for the voltage across the inductor, will be : Let $\pi=\frac{22}{7}$

(D) $23.1 \sin \left(49 \pi \mathrm{t}+60^{\circ}\right)$

(A) $1.47 \sin \left(49 \pi \mathrm{t}-30^{\circ}\right)$

(B) $1.47 \sin \left(49 \pi t+60^{\circ}\right)$

(C) $23.1 \sin \left(49 \pi \mathrm{t}-30^{\circ}\right)$

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Alternating Current

The equation of current in a purely inductive circuit is

5 sin (49 π t - 3 0^{\circ})

. If the inductance is 30 mH then the equation for the voltage across the inductor, will be :

Let

π = \frac{22}{7}

Answer: (D)

In a purely inductive circuit, the voltage amplitude is

V_{0} = I_{0} ω L

. Here

I_{0} = 5

ω = 49 π

rad/s, and

L = 30 \times 1 0^{- 3}

H. Thus

V_{0} = 5 \times 49 π \times 0.03 = 23.1

V. In an inductor, voltage leads current by

9 0^{\circ}

, so the phase becomes

- 3 0^{\circ} + 9 0^{\circ} = 6 0^{\circ}

. Therefore,

V = 23.1 sin (49 π t + 6 0^{\circ})

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option