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Current sensitivity of a moving coil galvanometeris 5 div/mA and its voltage sensitivity (angulardeflection per unit voltage applied) is 20 div/V. Theresistance of the galvanometer is
- a)40 Ω
- b)25 Ω
- c)250 Ω
- d)500 Ω
Correct answer is option 'C'. Can you explain this answer?
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To find the resistance of the galvanometer, we need to use Ohm's Law.
The current sensitivity of the galvanometer is given as 5 div/mA, which means that for every mA (milliampere) of current passing through the galvanometer, the deflection is 5 divisions. This can be converted to A (Amperes) by dividing by 1000:
Current sensitivity = 5 div/mA = 5 div / (0.001 A) = 5000 div/A
The voltage sensitivity of the galvanometer is given as 20 div/V, which means that for every volt applied to the galvanometer, the deflection is 20 divisions.
Voltage sensitivity = 20 div/V
We can combine these two sensitivities to find the overall sensitivity of the galvanometer:
Overall sensitivity = Current sensitivity / Voltage sensitivity
Overall sensitivity = (5000 div/A) / (20 div/V)
Overall sensitivity = 250 A/V
Now, we can use Ohm's Law to find the resistance of the galvanometer:
Resistance (R) = Voltage (V) / Current (I)
Resistance (R) = 1 V / (250 A/V)
Resistance (R) = 0.004 Ω
Therefore, the resistance of the galvanometer is 0.004 Ω.
The current sensitivity of the galvanometer is given as 5 div/mA, which means that for every mA (milliampere) of current passing through the galvanometer, the deflection is 5 divisions. This can be converted to A (Amperes) by dividing by 1000:
Current sensitivity = 5 div/mA = 5 div / (0.001 A) = 5000 div/A
The voltage sensitivity of the galvanometer is given as 20 div/V, which means that for every volt applied to the galvanometer, the deflection is 20 divisions.
Voltage sensitivity = 20 div/V
We can combine these two sensitivities to find the overall sensitivity of the galvanometer:
Overall sensitivity = Current sensitivity / Voltage sensitivity
Overall sensitivity = (5000 div/A) / (20 div/V)
Overall sensitivity = 250 A/V
Now, we can use Ohm's Law to find the resistance of the galvanometer:
Resistance (R) = Voltage (V) / Current (I)
Resistance (R) = 1 V / (250 A/V)
Resistance (R) = 0.004 Ω
Therefore, the resistance of the galvanometer is 0.004 Ω.
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Current sensitivity of a moving coil galvanometeris 5 div/mA and its voltage sensitivity (angulardeflection per unit voltage applied) is 20 div/V. Theresistance of the galvanometer isa)40 Ωb)25 Ωc)250 Ωd)500 ΩCorrect answer is option 'C'. Can you explain this answer?
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