The equation of current in a purely inductive circuit is 5 sin (49π t - 30°). If the inductance is 30 mH then the equation for the voltage across the inductor, will be: Let π = 22/7 (a) 1.47 sin (49π t - 30°) (b) 1.47 sin (49π t + 60°) (c) 23.1 sin (49π t - 30°) (d) 23.1 sin (49π t + 60°)

Answer: (d). XL = ω L = 49 × 22/7 × 30 × 10⁻³ = 154 × 0.03 = 4.62 Ω V₀ = I₀ XL = 5 × 4.62 = 23.1 V In a pure inductor the voltage leads the current by 90°: phase = -30° + 90° = +60°. So the answer is (d) 23.1 sin (49π t + 60°).

The equation of current in a purely inductive circuit is 5 sin (49π t… | JEE Main 2022 (28 Jul, Shift 1)

Q48JEE Main 2022 · 28 JUL · SHIFT 1Medium

Pure R, L and C Circuits

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

$\{$ Let $\pi = \frac{22}{7}\}$

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Answer: (d)

$X_L = \omega L = 49 \times \frac{22}{7} \times 30 \times 10^{-3} = 154 \times 0.03 = 4.62\ \Omega$
$V_0 = I_0 X_L = 5 \times 4.62 = 23.1$ V
In a pure inductor the voltage leads the current by $90^\circ$ : phase $= -30^\circ + 90^\circ = +60^\circ$ .
So the answer is (d) $23.1 \sin(49\pi t + 60^\circ)$ .

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