JEE Exam > JEE Questions > A sinusoidal voltage of peak value 283 V and ...
Start Learning for Free
A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5 Ω, L=25 mH and C=1000 µF. The total impedance, and phase difference between the voltage across the source and the current will respectively be :
- a)10 Ω and tan-1 (5/3)
- b)7Ω and 45o
- c)10 Ω and tan-1 (8/3)
- d)7Ω and tan-1(5/3)
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A sinusoidal voltage of peak value 283 V and angular frequency 320/s i...
NF, find:
a) The impedance of the circuit
b) The resonant frequency of the circuit
c) The quality factor of the circuit
d) The current in the circuit
e) The voltage across the capacitor
f) The voltage across the inductor
g) The voltage across the resistor
a) The impedance of the circuit can be calculated using the formula:
Z = √(R^2 + (ωL - 1/ωC)^2)
Substituting the given values:
Z = √(5^2 + (320*0.025 - 1/(320*0.001))^2) = 50.18 Ω
b) The resonant frequency of the circuit can be calculated using the formula:
ω0 = 1/√(LC)
Substituting the given values:
ω0 = 1/√(0.025*0.001) = 2000 rad/s
c) The quality factor of the circuit can be calculated using the formula:
Q = ω0L/R
Substituting the given values:
Q = (2000*0.025)/5 = 10
d) The current in the circuit can be calculated using Ohm's law:
I = V/Z
Substituting the given values:
I = 283/50.18 = 5.64 A
e) The voltage across the capacitor can be calculated using the formula:
Vc = I/(ωC)
Substituting the given values:
Vc = 5.64/(320*0.001) = 17.62 V
f) The voltage across the inductor can be calculated using the formula:
VL = I(ωL)
Substituting the given values:
VL = 5.64(320*0.025) = 45.31 V
g) The voltage across the resistor can be calculated using Ohm's law:
VR = IR
Substituting the given values:
VR = 5.64*5 = 28.2 V
a) The impedance of the circuit
b) The resonant frequency of the circuit
c) The quality factor of the circuit
d) The current in the circuit
e) The voltage across the capacitor
f) The voltage across the inductor
g) The voltage across the resistor
a) The impedance of the circuit can be calculated using the formula:
Z = √(R^2 + (ωL - 1/ωC)^2)
Substituting the given values:
Z = √(5^2 + (320*0.025 - 1/(320*0.001))^2) = 50.18 Ω
b) The resonant frequency of the circuit can be calculated using the formula:
ω0 = 1/√(LC)
Substituting the given values:
ω0 = 1/√(0.025*0.001) = 2000 rad/s
c) The quality factor of the circuit can be calculated using the formula:
Q = ω0L/R
Substituting the given values:
Q = (2000*0.025)/5 = 10
d) The current in the circuit can be calculated using Ohm's law:
I = V/Z
Substituting the given values:
I = 283/50.18 = 5.64 A
e) The voltage across the capacitor can be calculated using the formula:
Vc = I/(ωC)
Substituting the given values:
Vc = 5.64/(320*0.001) = 17.62 V
f) The voltage across the inductor can be calculated using the formula:
VL = I(ωL)
Substituting the given values:
VL = 5.64(320*0.025) = 45.31 V
g) The voltage across the resistor can be calculated using Ohm's law:
VR = IR
Substituting the given values:
VR = 5.64*5 = 28.2 V
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5 , L=25 mH and C=1000 µF. The total impedance, and phase difference between the voltage across the source and the current will respectively be :a)10 and tan-1 (5/3)b)7 and 45oc)10 and tan-1(8/3)d)7 and tan-1(5/3)Correct answer is option 'B'. Can you explain this answer?
Question Description
A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5 , L=25 mH and C=1000 µF. The total impedance, and phase difference between the voltage across the source and the current will respectively be :a)10 and tan-1 (5/3)b)7 and 45oc)10 and tan-1(8/3)d)7 and tan-1(5/3)Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5 , L=25 mH and C=1000 µF. The total impedance, and phase difference between the voltage across the source and the current will respectively be :a)10 and tan-1 (5/3)b)7 and 45oc)10 and tan-1(8/3)d)7 and tan-1(5/3)Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5 , L=25 mH and C=1000 µF. The total impedance, and phase difference between the voltage across the source and the current will respectively be :a)10 and tan-1 (5/3)b)7 and 45oc)10 and tan-1(8/3)d)7 and tan-1(5/3)Correct answer is option 'B'. Can you explain this answer?.
A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5 , L=25 mH and C=1000 µF. The total impedance, and phase difference between the voltage across the source and the current will respectively be :a)10 and tan-1 (5/3)b)7 and 45oc)10 and tan-1(8/3)d)7 and tan-1(5/3)Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5 , L=25 mH and C=1000 µF. The total impedance, and phase difference between the voltage across the source and the current will respectively be :a)10 and tan-1 (5/3)b)7 and 45oc)10 and tan-1(8/3)d)7 and tan-1(5/3)Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5 , L=25 mH and C=1000 µF. The total impedance, and phase difference between the voltage across the source and the current will respectively be :a)10 and tan-1 (5/3)b)7 and 45oc)10 and tan-1(8/3)d)7 and tan-1(5/3)Correct answer is option 'B'. Can you explain this answer?.
Solutions for A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5 , L=25 mH and C=1000 µF. The total impedance, and phase difference between the voltage across the source and the current will respectively be :a)10 and tan-1 (5/3)b)7 and 45oc)10 and tan-1(8/3)d)7 and tan-1(5/3)Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5 , L=25 mH and C=1000 µF. The total impedance, and phase difference between the voltage across the source and the current will respectively be :a)10 and tan-1 (5/3)b)7 and 45oc)10 and tan-1(8/3)d)7 and tan-1(5/3)Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5 , L=25 mH and C=1000 µF. The total impedance, and phase difference between the voltage across the source and the current will respectively be :a)10 and tan-1 (5/3)b)7 and 45oc)10 and tan-1(8/3)d)7 and tan-1(5/3)Correct answer is option 'B'. Can you explain this answer?, a detailed solution for A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5 , L=25 mH and C=1000 µF. The total impedance, and phase difference between the voltage across the source and the current will respectively be :a)10 and tan-1 (5/3)b)7 and 45oc)10 and tan-1(8/3)d)7 and tan-1(5/3)Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5 , L=25 mH and C=1000 µF. The total impedance, and phase difference between the voltage across the source and the current will respectively be :a)10 and tan-1 (5/3)b)7 and 45oc)10 and tan-1(8/3)d)7 and tan-1(5/3)Correct 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 angular frequency 320/s is applied to a series LCR circuit. Given that R=5 , L=25 mH and C=1000 µF. The total impedance, and phase difference between the voltage across the source and the current will respectively be :a)10 and tan-1 (5/3)b)7 and 45oc)10 and tan-1(8/3)d)7 and tan-1(5/3)Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup