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
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) = = 3.297 Omega.
-> Virtual Current (I) = V/X = 200/3.297 = 60.66 A
Given:
Voltage V
Frequency Hz
Inductance H
Resistance Ω
To Find:
(A)
In L-R circuit voltage leads current by angle,
Where
Ω
⇒
72.3°
(B)
The current in circuit is given by,
Where Circuit impedance
Since,
A
Therefore, the phase difference between current and emf is 72.3° and current in the circuit is 60.7 A