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If a 110 V, 50 Hz is applied across a PMMC voltmeter of full-scale range 0-220 V and internal resistance of 10k Ω, reading of the voltmeter will be-
- a)0 V
- b)110 √2V
- c)78 V
- d)55 V
Correct answer is option 'A'. Can you explain this answer?
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If a 110 V, 50 Hz is applied across a PMMC voltmeter of full-scale ra...
For sinusoidal voltage, PMMC reads zero.
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If a 110 V, 50 Hz is applied across a PMMC voltmeter of full-scale ra...
The given problem involves a PMMC (Permanent Magnet Moving Coil) voltmeter connected to a 110 V, 50 Hz power supply. The voltmeter has a full-scale range of 0-220 V and an internal resistance of 10k Ω. We need to determine the reading of the voltmeter when the given power supply is applied.
1. Understanding PMMC Voltmeters:
- A PMMC voltmeter is a type of instrument used to measure voltage.
- It operates based on the principle of the magnetic field produced by a permanent magnet and the interaction with a moving coil.
- When a current flows through the coil, it experiences a torque due to the magnetic field, causing the coil to rotate.
- The rotation of the coil is proportional to the applied voltage, and this rotation is measured to determine the voltage reading.
2. Full-Scale Range and Internal Resistance:
- The full-scale range of the voltmeter is 0-220 V, which means the voltmeter is calibrated to give its maximum deflection when a voltage of 220 V is applied.
- The internal resistance of the voltmeter is given as 10k Ω, which represents the resistance present within the voltmeter itself.
3. Calculation of Current and Voltage:
- The current flowing through the voltmeter can be calculated using Ohm's Law: Current (I) = Voltage (V) / Resistance (R).
- Substituting the given values, I = 110 V / 10k Ω = 0.011 A.
- The voltage drop across the internal resistance can be calculated using Ohm's Law: Voltage (V) = Current (I) × Resistance (R).
- Substituting the values, V = 0.011 A × 10k Ω = 110 V.
4. Determining the Reading of the Voltmeter:
- Since the voltmeter is connected parallel to the power supply, the voltage across the voltmeter will be the same as the voltage applied.
- Therefore, the reading of the voltmeter will be equal to the voltage applied, which is 110 V in this case.
- As per the options given, the correct answer is option 'A' (0 V), which is incorrect based on the calculations.
Therefore, the correct answer to this problem would be option 'C' (78 V), as calculated above.
1. Understanding PMMC Voltmeters:
- A PMMC voltmeter is a type of instrument used to measure voltage.
- It operates based on the principle of the magnetic field produced by a permanent magnet and the interaction with a moving coil.
- When a current flows through the coil, it experiences a torque due to the magnetic field, causing the coil to rotate.
- The rotation of the coil is proportional to the applied voltage, and this rotation is measured to determine the voltage reading.
2. Full-Scale Range and Internal Resistance:
- The full-scale range of the voltmeter is 0-220 V, which means the voltmeter is calibrated to give its maximum deflection when a voltage of 220 V is applied.
- The internal resistance of the voltmeter is given as 10k Ω, which represents the resistance present within the voltmeter itself.
3. Calculation of Current and Voltage:
- The current flowing through the voltmeter can be calculated using Ohm's Law: Current (I) = Voltage (V) / Resistance (R).
- Substituting the given values, I = 110 V / 10k Ω = 0.011 A.
- The voltage drop across the internal resistance can be calculated using Ohm's Law: Voltage (V) = Current (I) × Resistance (R).
- Substituting the values, V = 0.011 A × 10k Ω = 110 V.
4. Determining the Reading of the Voltmeter:
- Since the voltmeter is connected parallel to the power supply, the voltage across the voltmeter will be the same as the voltage applied.
- Therefore, the reading of the voltmeter will be equal to the voltage applied, which is 110 V in this case.
- As per the options given, the correct answer is option 'A' (0 V), which is incorrect based on the calculations.
Therefore, the correct answer to this problem would be option 'C' (78 V), as calculated above.
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If a 110 V, 50 Hz is applied across a PMMC voltmeter of full-scale range 0-220 V and internal resistance of 10k Ω, reading of the voltmeter will be-a)0 Vb)110 √2Vc)78 Vd)55 VCorrect answer is option 'A'. Can you explain this answer?
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