Question

An inductor with L=9.50 mH is connected across an ac source that has voltage amplitude 45.0 V. (a) What is the phase angle f for the source voltage relative to the current? Does the source voltage lag or lead the current? (b) What value for the frequency of the source results in a current amplitude of 3.90 A?

Short answer

(a) Phase angle, $\varphi =+\mathbf{90}\mathbf{°}$

(b) Frequency of the source, f =193 Hz

Step-by-step solution

Given Data

An inductor is a passive two-terminal device that stores energy in a magnetic field when current passes through it.

The time interval by which one wave leads or lags by another wave is called Phase angle or Phase difference. In Alternating current, we are mainly concerned with the phase difference between total voltage and total electric current.

The amplitude of voltage,V =45 V

The inductance of the coil, L =9.5 mH

Determination of Phase Angle

Inductors are made to react against change in current which causes them to lag behind the voltage. So, the phase angle will be in a positive sign and equal to 90°.

$\varphi =+90°$

Therefore, the phase angle for the source voltage relative to the current is + 90°

Determination of Source frequency

The amplitude of current,$l=3.9A$

Inductive reactance, $X_L=\frac{V_L}{I}=2\mathrm{\pi fL}$

Therefore,$f=\frac{V_L}{2\mathrm{\pi IL}}=\frac{45.V}{2\mathrm{\pi }*\left(9.5*{10}^{-3}\mathrm{H}\right)*\left(3.9\mathrm{A}\right)}=193Hz$

Thus, the frequency of the source is 193 Hertz.