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( Q No.A18052020).In a series RL circuit , the inductance being 20mH, the impedance is 17.85 ohms. The angle of lag of the input current from the applied voltage being 63.5 degrees find the values of angular frequency and the resistance of the circuit.?
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( Q No.A18052020).In a series RL circuit , the inductance being 20mH, ...
Answer:
Given parameters of the series RL circuit are:
- Inductance (L) = 20 mH
- Impedance (Z) = 17.85 Ω
- Angle of lag (θ) = 63.5°
We need to find the values of angular frequency (ω) and the resistance (R) of the circuit.
Formula:
- Impedance of RL circuit (Z) = sqrt(R^2 + (ωL)^2)
- Phase angle (θ) = tan^-1((ωL)/R)
Solution:
We know that the impedance of a series RL circuit is given by:
Z = sqrt(R^2 + (ωL)^2)
Squaring both sides of the equation, we get:
Z^2 = R^2 + (ωL)^2
Substituting the given values in the above equation, we get:
(17.85)^2 = R^2 + (ω x 20 x 10^-3)^2
R^2 + (ω x 20 x 10^-3)^2 = 319.0225
We also know that the phase angle (θ) of the circuit is given by:
θ = tan^-1((ωL)/R)
Substituting the given values, we get:
63.5° = tan^-1((ω x 20 x 10^-3)/R)
tan(63.5°) = (ω x 20 x 10^-3)/R
R x tan(63.5°) = ω x 20 x 10^-3
Squaring both sides of the above equation, we get:
R^2 x tan^2(63.5°) = (ω x 20 x 10^-3)^2
R^2 = ((ω x 20 x 10^-3)^2) / tan^2(63.5°)
Substituting the value of R^2 from the above equation in the equation of impedance, we get:
((ω x 20 x 10^-3)^2) / tan^2(63.5°) + (ω x 20 x 10^-3)^2 = 319.0225
Simplifying the above equation, we get a quadratic equation in ω^2:
(1 + tan^2(63.5°)) x (ω^2) = 319.0225 / (20 x 10^-3)^2
ω^2 = 684776.92
ω = sqrt(684776.92)
ω = 827.4 rad/s
Substituting the value of ω in the equation of resistance, we get:
R^2 = ((827.4 x 20 x 10^-3)^2) / tan^2(63.5°)
R = sqrt(((827.4 x 20 x 10^-3)^2) / tan^2(63.5°))
R = 56.63 Ω
Therefore, the values of angular frequency and the resistance of the circuit are:
- Angular frequency (ω) = 827.4 rad/s
- Resistance (R) = 56.63 Ω
Conclusion:
In this question, we have calculated the values of angular frequency
Given parameters of the series RL circuit are:
- Inductance (L) = 20 mH
- Impedance (Z) = 17.85 Ω
- Angle of lag (θ) = 63.5°
We need to find the values of angular frequency (ω) and the resistance (R) of the circuit.
Formula:
- Impedance of RL circuit (Z) = sqrt(R^2 + (ωL)^2)
- Phase angle (θ) = tan^-1((ωL)/R)
Solution:
We know that the impedance of a series RL circuit is given by:
Z = sqrt(R^2 + (ωL)^2)
Squaring both sides of the equation, we get:
Z^2 = R^2 + (ωL)^2
Substituting the given values in the above equation, we get:
(17.85)^2 = R^2 + (ω x 20 x 10^-3)^2
R^2 + (ω x 20 x 10^-3)^2 = 319.0225
We also know that the phase angle (θ) of the circuit is given by:
θ = tan^-1((ωL)/R)
Substituting the given values, we get:
63.5° = tan^-1((ω x 20 x 10^-3)/R)
tan(63.5°) = (ω x 20 x 10^-3)/R
R x tan(63.5°) = ω x 20 x 10^-3
Squaring both sides of the above equation, we get:
R^2 x tan^2(63.5°) = (ω x 20 x 10^-3)^2
R^2 = ((ω x 20 x 10^-3)^2) / tan^2(63.5°)
Substituting the value of R^2 from the above equation in the equation of impedance, we get:
((ω x 20 x 10^-3)^2) / tan^2(63.5°) + (ω x 20 x 10^-3)^2 = 319.0225
Simplifying the above equation, we get a quadratic equation in ω^2:
(1 + tan^2(63.5°)) x (ω^2) = 319.0225 / (20 x 10^-3)^2
ω^2 = 684776.92
ω = sqrt(684776.92)
ω = 827.4 rad/s
Substituting the value of ω in the equation of resistance, we get:
R^2 = ((827.4 x 20 x 10^-3)^2) / tan^2(63.5°)
R = sqrt(((827.4 x 20 x 10^-3)^2) / tan^2(63.5°))
R = 56.63 Ω
Therefore, the values of angular frequency and the resistance of the circuit are:
- Angular frequency (ω) = 827.4 rad/s
- Resistance (R) = 56.63 Ω
Conclusion:
In this question, we have calculated the values of angular frequency
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( Q No.A18052020).In a series RL circuit , the inductance being 20mH, ...
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( Q No.A18052020).In a series RL circuit , the inductance being 20mH, the impedance is 17.85 ohms. The angle of lag of the input current from the applied voltage being 63.5 degrees find the values of angular frequency and the resistance of the circuit.?
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( Q No.A18052020).In a series RL circuit , the inductance being 20mH, the impedance is 17.85 ohms. The angle of lag of the input current from the applied voltage being 63.5 degrees find the values of angular frequency and the resistance of the circuit.? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about ( Q No.A18052020).In a series RL circuit , the inductance being 20mH, the impedance is 17.85 ohms. The angle of lag of the input current from the applied voltage being 63.5 degrees find the values of angular frequency and the resistance of the circuit.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ( Q No.A18052020).In a series RL circuit , the inductance being 20mH, the impedance is 17.85 ohms. The angle of lag of the input current from the applied voltage being 63.5 degrees find the values of angular frequency and the resistance of the circuit.?.
( Q No.A18052020).In a series RL circuit , the inductance being 20mH, the impedance is 17.85 ohms. The angle of lag of the input current from the applied voltage being 63.5 degrees find the values of angular frequency and the resistance of the circuit.? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about ( Q No.A18052020).In a series RL circuit , the inductance being 20mH, the impedance is 17.85 ohms. The angle of lag of the input current from the applied voltage being 63.5 degrees find the values of angular frequency and the resistance of the circuit.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ( Q No.A18052020).In a series RL circuit , the inductance being 20mH, the impedance is 17.85 ohms. The angle of lag of the input current from the applied voltage being 63.5 degrees find the values of angular frequency and the resistance of the circuit.?.
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