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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?
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Verified 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
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Most Upvoted 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
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Community Answer
A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applie...
2 Ω, L = 0.1 H, and C = 10 μF. Calculate the impedance of the circuit and the current flowing through it.
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Read the following text and answer the following questions on the basis of the same:Tuning a radio set: In essence the simplest tuned radio frequency receiver is a simple crystal set. Desired frequency is tuned by a tuned coil / capacitor combination, and then the signal is presented to a simple crystal or diode detector where the amplitude modulated signal, is demodulated. This is then passed straight to the headphones or speaker. In radio set there is an LC oscillator comprising of a variable capacitor (or sometimes a variable coupling coil), with a knob on the front panel to tune the receiver. Capacitor used in old radio sets is gang capacitor. It consists of two sets of parallel circular plates one of which can rotate manually by means of a knob. The rotation causes overlapping areas of plates to change, thus changing its capacitance. Air gap between plates acts as dielectric. The capacitor has to be tuned in tandem corresponding to the frequency of a station so that the LC combination of the radio set resonates at the frequency of the desired station.When capacitive reactance (XC) is equal to the inductive reactance (XL), then the resonance occurs and the resonant frequency is given by ω0 = 1/√LCcurrent amplitude becomes maximum at the resonant frequency. It is important to note that resonance phenomenon is exhibited by a circuit only if both L and C are present in the circuit. Only then do the voltages across L and C cancel each other (both being out of phas e) and the Current amplitude is Vm/R the total source voltage appearing across R. This means that we cannot have resonance in a RL or RC circuit.Resonance may occur in

Read the following text and answer the following questions on the basis of the same: Tuning a radio set: In essence the simplest tuned radio frequency receiver is a simple crystal set. Desired frequency is tuned by a tuned coil / capacitor combination, and then the signal is presented to a simple crystal or diode detector where the amplitude modulated signal, is demodulated. This is then passed straight to the headphones or speaker. In radio set there is an LC oscillator comprising of a variable capacitor (or sometimes a variable coupling coil), with a knob on the front panel to tune the receiver. Capacitor used in old radio sets is gang capacitor. It consists of two sets of parallel circular plates one of which can rotate manually by means of a knob. The rotation causes overlapping areas of plates to change, thus changing its capacitance. Air gap between plates acts as dielectric. The capacitor has to be tuned in tandem corresponding to the frequency of a station so that the LC combination of the radio set resonates at the frequency of the desired station.When capacitive reactance (XC) is equal to the inductive reactance (XL), then the resonance occurs and the resonant frequency is given by ω0 = 1/√LCcurrent amplitude becomes maximum at the resonant frequency. It is important to note that resonance phenomenon is exhibited by a circuit only if both L and C are present in the circuit. Only then do the voltages across L and C cancel each other (both being out of phas e) and the Current amplitude is Vm/R the total source voltage appearing across R. This means that we cannot have resonance in a RL or RC circuit.Resonance frequency is equal to

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?
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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 2024 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 2024 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. Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.
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