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A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH, and C = 796 μF. Power dissipated in the circuit; and the power factor are
- a)4000 W, 0.4
- b)4800 W, 0.6
- c)4400 W, 0.6
- d)3800 W, 0.6
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applie...
Angular frequency of the ac signal w=2πν
∴ w=2π(50)=100π
Capacitive reactance Xc=1/wC
∴ Xc=1/(100π×786×10−6)=4Ω
Inductive reactance XL=wL
∴ XL=100π×(25.48×10−3)=8Ω
Impedance of the circuit Z=√[R2+(XL−Xc)2]
∴ Z=√[32+(8−4)2]=5Ω
Phase difference ϕ=tan−1[(XL−Xc)R]
Or ϕ=tan−1((8−4)/3)=tan−1(4/3)
⟹ ϕ=53.13o
Power factor cosϕ=cos53.13o=0.6
Power dissipated in the circuit
P=Iv2R
Now, Iv=I0/√2=E0/√2Z=283/(1.414×5)=40A
∴P=Iv2R=(40)2×3=4800 watt
∴ w=2π(50)=100π
Capacitive reactance Xc=1/wC
∴ Xc=1/(100π×786×10−6)=4Ω
Inductive reactance XL=wL
∴ XL=100π×(25.48×10−3)=8Ω
Impedance of the circuit Z=√[R2+(XL−Xc)2]
∴ Z=√[32+(8−4)2]=5Ω
Phase difference ϕ=tan−1[(XL−Xc)R]
Or ϕ=tan−1((8−4)/3)=tan−1(4/3)
⟹ ϕ=53.13o
Power factor cosϕ=cos53.13o=0.6
Power dissipated in the circuit
P=Iv2R
Now, Iv=I0/√2=E0/√2Z=283/(1.414×5)=40A
∴P=Iv2R=(40)2×3=4800 watt
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Community Answer
A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applie...
To find the impedance of circuit we first calculate XL and Xc XL=2pievl=2×3.14×50×25.4 8×10^-3=8ohm
Xc=1/2pievc=1/2×3.14×50×796×10^-6=4ohm
Z=under root R^2(XL-Xc)^2=underoot3^2(8-4)^2=5ohm
phase difference for=tan^-1(Xc-XL)/R=
tan^-1 4-8/3=-53.1degree
since fir is(-ve) the current in the circuit lags the voltage across the source.
the power dissipated in the circuit is
P=I2R
I=Im/root 2=(1/root2)(283/5)=40A
P=(40)^2×3=4800
Power factor=cosfie=cos 53.1degree=0.6
Xc=1/2pievc=1/2×3.14×50×796×10^-6=4ohm
Z=under root R^2(XL-Xc)^2=underoot3^2(8-4)^2=5ohm
phase difference for=tan^-1(Xc-XL)/R=
tan^-1 4-8/3=-53.1degree
since fir is(-ve) the current in the circuit lags the voltage across the source.
the power dissipated in the circuit is
P=I2R
I=Im/root 2=(1/root2)(283/5)=40A
P=(40)^2×3=4800
Power factor=cosfie=cos 53.1degree=0.6
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A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω, L = 25.48 mH, and C = 796μF. Power dissipated in the circuit; and the power factor area)4000 W, 0.4b)4800 W, 0.6c)4400 W, 0.6d)3800 W, 0.6Correct answer is option 'B'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω, L = 25.48 mH, and C = 796μF. Power dissipated in the circuit; and the power factor area)4000 W, 0.4b)4800 W, 0.6c)4400 W, 0.6d)3800 W, 0.6Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω, L = 25.48 mH, and C = 796μF. Power dissipated in the circuit; and the power factor area)4000 W, 0.4b)4800 W, 0.6c)4400 W, 0.6d)3800 W, 0.6Correct answer is option 'B'. Can you explain this answer?.
A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω, L = 25.48 mH, and C = 796μF. Power dissipated in the circuit; and the power factor area)4000 W, 0.4b)4800 W, 0.6c)4400 W, 0.6d)3800 W, 0.6Correct answer is option 'B'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω, L = 25.48 mH, and C = 796μF. Power dissipated in the circuit; and the power factor area)4000 W, 0.4b)4800 W, 0.6c)4400 W, 0.6d)3800 W, 0.6Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω, L = 25.48 mH, and C = 796μF. Power dissipated in the circuit; and the power factor area)4000 W, 0.4b)4800 W, 0.6c)4400 W, 0.6d)3800 W, 0.6Correct answer is option 'B'. Can you explain this answer?.
Solutions for A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω, L = 25.48 mH, and C = 796μF. Power dissipated in the circuit; and the power factor area)4000 W, 0.4b)4800 W, 0.6c)4400 W, 0.6d)3800 W, 0.6Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω, L = 25.48 mH, and C = 796μF. Power dissipated in the circuit; and the power factor area)4000 W, 0.4b)4800 W, 0.6c)4400 W, 0.6d)3800 W, 0.6Correct answer is option 'B'. Can you explain this answer?, a detailed solution for A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω, L = 25.48 mH, and C = 796μF. Power dissipated in the circuit; and the power factor area)4000 W, 0.4b)4800 W, 0.6c)4400 W, 0.6d)3800 W, 0.6Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω, L = 25.48 mH, and C = 796μF. Power dissipated in the circuit; and the power factor area)4000 W, 0.4b)4800 W, 0.6c)4400 W, 0.6d)3800 W, 0.6Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω, L = 25.48 mH, and C = 796μF. Power dissipated in the circuit; and the power factor area)4000 W, 0.4b)4800 W, 0.6c)4400 W, 0.6d)3800 W, 0.6Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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