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A student in a lab took a coil and connected it to a 12 V DC source. He measures the steady state current in the circuit to be 4A. He then replaced the 12 V DC source by a 12 V, (w = 50 rad/s) AC source and observes that the reading in the AC ammeter is 2.4 A. He then decides to connect a 2500 mF capacitor in series with the coil and calculate the average power developed in the circuit. Further he also decides to study the variation in current in the circuit (with the capacitor and the battery in series).
Based on the readings taken by the student answer the following questions.
The power developed in the circuit when the capacitor of 2500 mF is connected in series with the coil is
  • a)
     28.8 W
  • b)
    23.04 W
  • c)
    17.28 W
  • d)
    9.6 W
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A student in a lab took a coil and connected it to a 12 V DC source. H...
VDC=IDCR
∴R=VDC/IDC=12/4=3Ω
IAC=VAC/Z=VAC√(R2+XL2)
2.4=12√ [(3)2+XL2]
solving this equation, we get,
XL=4Ω
XC=1/ωC=1/(50×2500×10−6)
=8Ω
Z=√[R2+(XC−XL)2] =5Ω
∴I=VDC/Z=12/5=2.4A=Irms
P=Irms2R=(2.4)2(3)
=17.28W
At given frequency XC>XL if omega is further decreased XC will increase (as XC∝1/ω) and XL will increase (as XL∝ω).
Therefore XC−XL and hence Z will increase. So, current will decrease.
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Most Upvoted Answer
A student in a lab took a coil and connected it to a 12 V DC source. H...
Understanding the Circuit Components
The student initially works with a coil connected to a 12 V DC source, measuring a steady current of 4 A. When switched to a 12 V AC source at 50 rad/s, the current drops to 2.4 A. The introduction of a 2500 mF capacitor in series with the coil affects the circuit dynamics.
Calculating Impedance and Current
- The impedance (Z) of the circuit can be calculated since the current in AC differs from DC.
- The resistance (R) of the coil can be derived from DC conditions:
R = V/I = 12V / 4A = 3Ω.
- The reactance (Xc) of the capacitor is:
Xc = 1 / (ωC) = 1 / (50 * 2500 * 10^-3) = 0.008Ω.
- The total impedance (Z) in the series circuit is:
Z = √(R^2 + Xc^2) = √(3^2 + 0.008^2) ≈ 3Ω.
Average Power Calculation
- The average power (P) in an AC circuit can be calculated using the formula:
P = I^2 * R.
- Now, substituting the AC current (I = 2.4 A) and the resistance (R = 3Ω):
P = (2.4 A)^2 * 3Ω = 5.76 * 3 = 17.28 W.
Conclusion
Thus, the average power developed in the circuit when the 2500 mF capacitor is connected in series with the coil is 17.28 W, confirming that option 'C' is indeed correct.
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A student in a lab took a coil and connected it to a 12 V DC source. He measures the steady state current in the circuit to be 4A. He then replaced the 12 V DC source by a 12 V, (w = 50 rad/s) AC source and observes that the reading in the AC ammeter is 2.4 A. He then decides to connect a 2500 mF capacitor in series with the coil and calculate the average power developed in the circuit. Further he also decides to study the variation in current in the circuit (with the capacitor and the battery in series).Based on the readings taken by the student answer the following questions.The power developed in the circuit when the capacitor of 2500 mF is connected in series with the coil isa)28.8 Wb)23.04 Wc)17.28 Wd)9.6 WCorrect answer is option 'C'. Can you explain this answer?
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A student in a lab took a coil and connected it to a 12 V DC source. He measures the steady state current in the circuit to be 4A. He then replaced the 12 V DC source by a 12 V, (w = 50 rad/s) AC source and observes that the reading in the AC ammeter is 2.4 A. He then decides to connect a 2500 mF capacitor in series with the coil and calculate the average power developed in the circuit. Further he also decides to study the variation in current in the circuit (with the capacitor and the battery in series).Based on the readings taken by the student answer the following questions.The power developed in the circuit when the capacitor of 2500 mF is connected in series with the coil isa)28.8 Wb)23.04 Wc)17.28 Wd)9.6 WCorrect answer is option 'C'. 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 student in a lab took a coil and connected it to a 12 V DC source. He measures the steady state current in the circuit to be 4A. He then replaced the 12 V DC source by a 12 V, (w = 50 rad/s) AC source and observes that the reading in the AC ammeter is 2.4 A. He then decides to connect a 2500 mF capacitor in series with the coil and calculate the average power developed in the circuit. Further he also decides to study the variation in current in the circuit (with the capacitor and the battery in series).Based on the readings taken by the student answer the following questions.The power developed in the circuit when the capacitor of 2500 mF is connected in series with the coil isa)28.8 Wb)23.04 Wc)17.28 Wd)9.6 WCorrect answer is option 'C'. 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 student in a lab took a coil and connected it to a 12 V DC source. He measures the steady state current in the circuit to be 4A. He then replaced the 12 V DC source by a 12 V, (w = 50 rad/s) AC source and observes that the reading in the AC ammeter is 2.4 A. He then decides to connect a 2500 mF capacitor in series with the coil and calculate the average power developed in the circuit. Further he also decides to study the variation in current in the circuit (with the capacitor and the battery in series).Based on the readings taken by the student answer the following questions.The power developed in the circuit when the capacitor of 2500 mF is connected in series with the coil isa)28.8 Wb)23.04 Wc)17.28 Wd)9.6 WCorrect answer is option 'C'. Can you explain this answer?.
Solutions for A student in a lab took a coil and connected it to a 12 V DC source. He measures the steady state current in the circuit to be 4A. He then replaced the 12 V DC source by a 12 V, (w = 50 rad/s) AC source and observes that the reading in the AC ammeter is 2.4 A. He then decides to connect a 2500 mF capacitor in series with the coil and calculate the average power developed in the circuit. Further he also decides to study the variation in current in the circuit (with the capacitor and the battery in series).Based on the readings taken by the student answer the following questions.The power developed in the circuit when the capacitor of 2500 mF is connected in series with the coil isa)28.8 Wb)23.04 Wc)17.28 Wd)9.6 WCorrect answer is option 'C'. 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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