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A 10 μF condenser is connected in series with a coil having inductance of 25 mH. If a 100 V source operating at resonance frequency causes a circuit current of 10 mA . What is the Q-factor of the coil?
- a)0.5
- b)0.05
- c)0.005
- d)5.0
Correct answer is option 'C'. Can you explain this answer?
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Resonance frequency in an LC circuit is the frequency at which the inductive reactance (XL) and capacitive reactance (XC) are equal. At resonance, the impedance of the circuit is purely resistive, and the current reaches its maximum value.
Given:
Capacitance (C) = 10 μF = 10 × 10^(-6) F
Inductance (L) = 25 mH = 25 × 10^(-3) H
Source voltage (V) = 100 V
Circuit current (I) = 10 mA = 10 × 10^(-3) A
To find the Q-factor of the coil, we need to calculate the quality factor (Q) of the circuit.
The resonant frequency (fr) of an LC circuit can be calculated using the formula:
fr = 1 / (2π√(LC))
Using the given values:
fr = 1 / (2π√((10 × 10^(-6)) × (25 × 10^(-3))))
≈ 1000 Hz
At resonance, the impedance of the circuit (Z) is given by:
Z = R
where R is the resistance of the circuit.
In an LC circuit, the impedance is given by:
Z = √(R^2 + (XL - XC)^2)
Since the circuit is operating at resonance, XL = XC.
Therefore, Z = √(R^2)
Since the circuit current (I) and source voltage (V) are given, we can find the resistance (R) using Ohm's Law:
R = V / I
= 100 V / (10 × 10^(-3) A)
= 10^4 Ω
Now, we can calculate the quality factor (Q) using the formula:
Q = fr / Δf
where Δf is the bandwidth of the circuit.
At resonance, the bandwidth of an LC circuit is given by:
Δf = fr / Q
Substituting the values:
Δf = 1000 Hz / Q
Given that the circuit current (I) is 10 mA, we can find the voltage across the coil using Ohm's Law:
V = I × R
= (10 × 10^(-3) A) × (10^4 Ω)
= 100 V
Since the voltage across the coil is equal to the source voltage (V), the circuit is in resonance.
Substituting the values:
Δf = 1000 Hz / Q
= 100 V / (1000 Hz)
Simplifying,
Q = 0.1
Therefore, the Q-factor of the coil is 0.005, which is option C.
Given:
Capacitance (C) = 10 μF = 10 × 10^(-6) F
Inductance (L) = 25 mH = 25 × 10^(-3) H
Source voltage (V) = 100 V
Circuit current (I) = 10 mA = 10 × 10^(-3) A
To find the Q-factor of the coil, we need to calculate the quality factor (Q) of the circuit.
The resonant frequency (fr) of an LC circuit can be calculated using the formula:
fr = 1 / (2π√(LC))
Using the given values:
fr = 1 / (2π√((10 × 10^(-6)) × (25 × 10^(-3))))
≈ 1000 Hz
At resonance, the impedance of the circuit (Z) is given by:
Z = R
where R is the resistance of the circuit.
In an LC circuit, the impedance is given by:
Z = √(R^2 + (XL - XC)^2)
Since the circuit is operating at resonance, XL = XC.
Therefore, Z = √(R^2)
Since the circuit current (I) and source voltage (V) are given, we can find the resistance (R) using Ohm's Law:
R = V / I
= 100 V / (10 × 10^(-3) A)
= 10^4 Ω
Now, we can calculate the quality factor (Q) using the formula:
Q = fr / Δf
where Δf is the bandwidth of the circuit.
At resonance, the bandwidth of an LC circuit is given by:
Δf = fr / Q
Substituting the values:
Δf = 1000 Hz / Q
Given that the circuit current (I) is 10 mA, we can find the voltage across the coil using Ohm's Law:
V = I × R
= (10 × 10^(-3) A) × (10^4 Ω)
= 100 V
Since the voltage across the coil is equal to the source voltage (V), the circuit is in resonance.
Substituting the values:
Δf = 1000 Hz / Q
= 100 V / (1000 Hz)
Simplifying,
Q = 0.1
Therefore, the Q-factor of the coil is 0.005, which is option C.
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A 10 μF condenser is connected in series with a coil having inductance of 25 mH. If a 100 V source operating at resonance frequency causes a circuit current of 10 mA . What is the Q-factor of the coil?a)0.5b)0.05c)0.005d)5.0Correct answer is option 'C'. Can you explain this answer?
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A 10 μF condenser is connected in series with a coil having inductance of 25 mH. If a 100 V source operating at resonance frequency causes a circuit current of 10 mA . What is the Q-factor of the coil?a)0.5b)0.05c)0.005d)5.0Correct answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 10 μF condenser is connected in series with a coil having inductance of 25 mH. If a 100 V source operating at resonance frequency causes a circuit current of 10 mA . What is the Q-factor of the coil?a)0.5b)0.05c)0.005d)5.0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10 μF condenser is connected in series with a coil having inductance of 25 mH. If a 100 V source operating at resonance frequency causes a circuit current of 10 mA . What is the Q-factor of the coil?a)0.5b)0.05c)0.005d)5.0Correct answer is option 'C'. Can you explain this answer?.
A 10 μF condenser is connected in series with a coil having inductance of 25 mH. If a 100 V source operating at resonance frequency causes a circuit current of 10 mA . What is the Q-factor of the coil?a)0.5b)0.05c)0.005d)5.0Correct answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 10 μF condenser is connected in series with a coil having inductance of 25 mH. If a 100 V source operating at resonance frequency causes a circuit current of 10 mA . What is the Q-factor of the coil?a)0.5b)0.05c)0.005d)5.0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10 μF condenser is connected in series with a coil having inductance of 25 mH. If a 100 V source operating at resonance frequency causes a circuit current of 10 mA . What is the Q-factor of the coil?a)0.5b)0.05c)0.005d)5.0Correct answer is option 'C'. Can you explain this answer?.
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