Question

An alternating emf 200 virtual volts at 50 Hz is connected to a circuit resistance 1omega and inductance 0.01 H. What is the phase difference between the current and the emf in the circuit? Also, find the virtual current in the circuit.

Answer 1

The phase difference between the current and the emf in the circuit is = 72.34°

The Virtual Current in the circuit is = 60.66 A

->   Given: Voltage (V) = 200 V; Frequency (v) = 50 Hz; Resistance = 1 Omega; Inductance (L) = 0.01 H.

->   Angular frequency (w) = 2*pi*v = 100pi.

->   Reactance of Inductor (Xl) = wL = 100*pi*0.01 = pi = 3.14 Omega.

->   Phase difference (∅) = tan^-1(Xl/R) = tan^-1(pi) = 72.34°

->   Total Reactance (X) = $\sqrt{R^{2} + Xl^{2} }$ = 3.297 Omega.

->   Virtual Current (I) = V/X = 200/3.297 = 60.66 A

Answer 2

Given:

Voltage $V_{rms} = 200$ V

Frequency $f = 50$ Hz

Inductance $L = 0.01$ H

Resistance $R = 1$ Ω

To Find:

(A)

In L-R circuit voltage leads current by angle,

  $\phi = \tan^{-1} (\frac{X_{L} }{R} )$

Where $X_{L} = \omega L = 2\pi fL$

$X_{L} = 2\pi \times 50 \times 0.01$

$X_{L} = \pi$ Ω

⇒ $\phi = \tan^{-1} (\pi )$

    $\phi =$ 72.3°

(B)

The current in circuit is given by,

 $i _{rms} = \frac{V_{rms} }{Z}$

Where $Z =$ Circuit impedance

  $Z = \sqrt{R^{2} + X_{L} ^{2} }$

  $Z = \sqrt{1^{2} + \pi ^{2} }$

Since,  $i_{rms} = \frac{200}{{1^{2} + \pi ^{2} }}$

 $i_{rms} = 60.7$A

Therefore, the phase difference between current and emf is 72.3° and current in the circuit is 60.7 A