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A parallel plate capacitor made of circular plates each of radius R = 5 cm has a capacitance C = 50 pF. The capacitor is connected to a 220 V AC
supply with an angular frequency of 50 rad s–1. Then the RMS value of displacement current across the plates is (in μA)
1)0.275
2)0.55
3)1.1
4)2.2?
supply with an angular frequency of 50 rad s–1. Then the RMS value of displacement current across the plates is (in μA)
1)0.275
2)0.55
3)1.1
4)2.2?
Most Upvoted Answer
A parallel plate capacitor made of circular plates each of radius R = ...
Capacitance and AC Supply
The given parallel plate capacitor has a capacitance (C) of 50 pF and is connected to a 220 V AC supply with an angular frequency (ω) of 50 rad/s.
Displacement Current
In AC circuits, the displacement current (I_d) can be calculated using the formula:
I_d = C * (dV/dt)
Where:
- C = Capacitance
- dV/dt = The time derivative of the voltage across the capacitor.
Calculating dV/dt
Since the voltage is AC, it can be expressed as:
V(t) = V_m * sin(ωt)
Where:
- V_m = 220 V (maximum voltage)
- ω = 50 rad/s
The derivative of V(t) is:
dV/dt = V_m * ω * cos(ωt)
For RMS calculations, we consider the average value over a complete cycle. The RMS value of the displacement current can be given by:
I_d(rms) = C * V_m * ω / √2
Substituting Values
Now substituting the values:
- C = 50 pF = 50 x 10^-12 F
- V_m = 220 V
- ω = 50 rad/s
Calculating I_d(rms):
I_d(rms) = (50 x 10^-12) * 220 * 50 / √2
This simplifies to:
I_d(rms) = (50 * 220 * 50) / (√2 * 10^12)
Calculating numerically:
I_d(rms) = (5500000) / (1.414 * 10^12) = 0.275 μA
Conclusion
Thus, the RMS value of the displacement current across the plates is 0.275 μA.
The correct option is:
- Option 1: 0.275 μA
The given parallel plate capacitor has a capacitance (C) of 50 pF and is connected to a 220 V AC supply with an angular frequency (ω) of 50 rad/s.
Displacement Current
In AC circuits, the displacement current (I_d) can be calculated using the formula:
I_d = C * (dV/dt)
Where:
- C = Capacitance
- dV/dt = The time derivative of the voltage across the capacitor.
Calculating dV/dt
Since the voltage is AC, it can be expressed as:
V(t) = V_m * sin(ωt)
Where:
- V_m = 220 V (maximum voltage)
- ω = 50 rad/s
The derivative of V(t) is:
dV/dt = V_m * ω * cos(ωt)
For RMS calculations, we consider the average value over a complete cycle. The RMS value of the displacement current can be given by:
I_d(rms) = C * V_m * ω / √2
Substituting Values
Now substituting the values:
- C = 50 pF = 50 x 10^-12 F
- V_m = 220 V
- ω = 50 rad/s
Calculating I_d(rms):
I_d(rms) = (50 x 10^-12) * 220 * 50 / √2
This simplifies to:
I_d(rms) = (50 * 220 * 50) / (√2 * 10^12)
Calculating numerically:
I_d(rms) = (5500000) / (1.414 * 10^12) = 0.275 μA
Conclusion
Thus, the RMS value of the displacement current across the plates is 0.275 μA.
The correct option is:
- Option 1: 0.275 μA
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A parallel plate capacitor made of circular plates each of radius R = 5 cm has a capacitance C = 50 pF. The capacitor is connected to a 220 V AC supply with an angular frequency of 50 rad s–1. Then the RMS value of displacement current across the plates is (in μA) 1)0.275 2)0.55 3)1.1 4)2.2?
Question Description
A parallel plate capacitor made of circular plates each of radius R = 5 cm has a capacitance C = 50 pF. The capacitor is connected to a 220 V AC supply with an angular frequency of 50 rad s–1. Then the RMS value of displacement current across the plates is (in μA) 1)0.275 2)0.55 3)1.1 4)2.2? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A parallel plate capacitor made of circular plates each of radius R = 5 cm has a capacitance C = 50 pF. The capacitor is connected to a 220 V AC supply with an angular frequency of 50 rad s–1. Then the RMS value of displacement current across the plates is (in μA) 1)0.275 2)0.55 3)1.1 4)2.2? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate capacitor made of circular plates each of radius R = 5 cm has a capacitance C = 50 pF. The capacitor is connected to a 220 V AC supply with an angular frequency of 50 rad s–1. Then the RMS value of displacement current across the plates is (in μA) 1)0.275 2)0.55 3)1.1 4)2.2?.
A parallel plate capacitor made of circular plates each of radius R = 5 cm has a capacitance C = 50 pF. The capacitor is connected to a 220 V AC supply with an angular frequency of 50 rad s–1. Then the RMS value of displacement current across the plates is (in μA) 1)0.275 2)0.55 3)1.1 4)2.2? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A parallel plate capacitor made of circular plates each of radius R = 5 cm has a capacitance C = 50 pF. The capacitor is connected to a 220 V AC supply with an angular frequency of 50 rad s–1. Then the RMS value of displacement current across the plates is (in μA) 1)0.275 2)0.55 3)1.1 4)2.2? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate capacitor made of circular plates each of radius R = 5 cm has a capacitance C = 50 pF. The capacitor is connected to a 220 V AC supply with an angular frequency of 50 rad s–1. Then the RMS value of displacement current across the plates is (in μA) 1)0.275 2)0.55 3)1.1 4)2.2?.
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