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

Question

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

Tardigrade Admin · Accepted Answer

v0=i0 xL =i0(w L) =(5)(49 π)(30 × 10-3) =23.1 Voltage will lead current by 90°. ∴ V =23.1 sin (49 π t +60°)

Q. The equation of current in a purely inductive circuit is $5 sin (49 π t - 3 0^{\circ})$ . If the inductance is $30 m H$ then the equation for the voltage across the inductor, will be:
${Let π = \frac{22}{7}}$

$1.47 sin (49 π t - 3 0^{\circ})$

$1.47 sin (49 π t + 6 0^{\circ})$

$23.1 sin (49 π t - 3 0^{\circ})$

$23.1 sin (49 π t + 6 0^{\circ})$

$v_{0} = i_{0} x_{L}$
$= i_{0} (w L)$
$= (5) (49 π) (30 \times 1 0^{- 3})$
$= 23.1$
Voltage will lead current by $9 0^{\circ}$ .
$∴ V = 23.1 sin (49 π t + 6 0^{\circ})$

Q. The equation of current in a purely inductive circuit is 5sin(49πt−30∘). If the inductance is 30mH then the equation for the voltage across the inductor, will be: { Let π=722​}

1.47sin(49πt−30∘)

1.47sin(49πt+60∘)

23.1sin(49πt−30∘)

23.1sin(49πt+60∘)