An inductor is connected to an AC source, When compared to voltage, the current in the lead wires

Question

An inductor is connected to an AC source, When compared to voltage, the current in the lead wires (A) is ahead in phase by π (B) lags in phase by π (C) is ahead in phase by (π/2) (D) lags in phase by (π/2)

Tardigrade Admin · Accepted Answer

In a purely inductive circuit, current is

i = i_{0} sin (ω t - \frac{π}{2})

Which shows that the current lags behind the emf by a phase angle of

\frac{π}{2}

or

9 0^{\circ}

or the emf leads the current by a phase angle of

\frac{π}{2}

or

9 0^{\circ}

Q. An inductor is connected to an AC source, When compared to voltage, the current in the lead wires

is ahead in phase by $π$

lags in phase by $π$

is ahead in phase by $\frac{π}{2}$

lags in phase by $\frac{π}{2}$

In a purely inductive circuit, current is
$i = i_{0} sin (ω t - \frac{π}{2})$
Which shows that the current lags behind the emf by a phase angle of
$\frac{π}{2}$ or $9 0^{\circ}$
or the emf leads the current by a phase angle of
$\frac{π}{2}$ or $9 0^{\circ}$

Q. An inductor is connected to an AC source, When compared to voltage, the current in the lead wires

is ahead in phase by π

lags in phase by π

is ahead in phase by 2π​

lags in phase by 2π​

In a purely inductive circuit, current is i=i0​sin(ωt−2π​) Which shows that the current lags behind the emf by a phase angle of 2π​ or 90∘ or the emf leads the current by a phase angle of 2π​ or 90∘

is ahead in phase by $π$

lags in phase by $π$

is ahead in phase by $\frac{π}{2}$

lags in phase by $\frac{π}{2}$

In a purely inductive circuit, current is
$i = i_{0} sin (ω t - \frac{π}{2})$
Which shows that the current lags behind the emf by a phase angle of
$\frac{π}{2}$ or $9 0^{\circ}$
or the emf leads the current by a phase angle of
$\frac{π}{2}$ or $9 0^{\circ}$