Physics, asked by theRockstar5096, 9 months ago

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.

Answers

Answered by qwchair
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

Answered by minku8906
0

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} = 60.7A

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

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