The relation between voltage sensitivity\[({\sigma _v})\]and current sensitivity \[({\sigma _i})\]of a moving coil galvanometer is: (resistance of galvanometer is G).
A. \[\dfrac{{{\sigma _i}}}{G} = {\sigma _v}\]
B. \[\dfrac{{{\sigma _v}}}{G} = {\sigma _i}\]
C. \[\dfrac{G}{{{\sigma _v}}} = {\sigma _i}\]
D. None of the above
Answer
299.1k+ views
Hint:Current sensitivity of a moving coil galvanometer is defined as deflection produced in its coil per unit flow of current. The deflection produced in the coil per unit potential difference across it is called voltage sensitivity. Resistance is the ratio of potential difference and current as given by Ohm’s law.
Formula used:
Ohm’s Law, \[V = iR\]
where V = potential difference across the resistor, i = current across the resistor and R = resistance.
Current sensitivity = \[{\sigma _i} = \dfrac{\theta }{i} = \dfrac{{NBA}}{K}\]
Voltage sensitivity = \[{\sigma _v} = \dfrac{\theta }{V} = \dfrac{\theta }{{iG}} = \dfrac{{NBA}}{{KG}}\]
where\[\theta \]= deflection produced in the galvanometer (in radians),
i = current flowing through the galvanometer coil
V= potential difference across the galvanometer coil
B = magnetic field strength about the coil
A = area of the coil
N = number of turns
K = restoring torque
Complete step by step solution:
Given: Current sensitivity of a moving coil galvanometer = \[{\sigma _i}\]
Voltage sensitivity = \[{\sigma _v}\]
Resistance of the galvanometer = G
We know that, Current sensitivity = deflection per unit current = \[{\sigma _i} = \dfrac{\theta }{i}\] -- (1)
Voltage sensitivity = deflection per unit potential difference = \[{\sigma _v} = \dfrac{\theta }{V} = \dfrac{\theta }{{iG}}\]-- (2)
Substituting (1) in (2)
We get, \[{\sigma _v} = \dfrac{{{\sigma _i}}}{G}\]
Hence option (A) is the correct answer.
Note: Although it seems that the voltage sensitivity is directly proportional to current sensitivity, however, the voltage sensitivity in the moving coil galvanometer may not always rise with increasing current sensitivity. This is because the current sensitivity is directly proportional to the number of turns of the coil. Increasing the number of turns of the coil also increases the resistance of the coil which can negatively affect the voltage sensitivity.
Formula used:
Ohm’s Law, \[V = iR\]
where V = potential difference across the resistor, i = current across the resistor and R = resistance.
Current sensitivity = \[{\sigma _i} = \dfrac{\theta }{i} = \dfrac{{NBA}}{K}\]
Voltage sensitivity = \[{\sigma _v} = \dfrac{\theta }{V} = \dfrac{\theta }{{iG}} = \dfrac{{NBA}}{{KG}}\]
where\[\theta \]= deflection produced in the galvanometer (in radians),
i = current flowing through the galvanometer coil
V= potential difference across the galvanometer coil
B = magnetic field strength about the coil
A = area of the coil
N = number of turns
K = restoring torque
Complete step by step solution:
Given: Current sensitivity of a moving coil galvanometer = \[{\sigma _i}\]
Voltage sensitivity = \[{\sigma _v}\]
Resistance of the galvanometer = G
We know that, Current sensitivity = deflection per unit current = \[{\sigma _i} = \dfrac{\theta }{i}\] -- (1)
Voltage sensitivity = deflection per unit potential difference = \[{\sigma _v} = \dfrac{\theta }{V} = \dfrac{\theta }{{iG}}\]-- (2)
Substituting (1) in (2)
We get, \[{\sigma _v} = \dfrac{{{\sigma _i}}}{G}\]
Hence option (A) is the correct answer.
Note: Although it seems that the voltage sensitivity is directly proportional to current sensitivity, however, the voltage sensitivity in the moving coil galvanometer may not always rise with increasing current sensitivity. This is because the current sensitivity is directly proportional to the number of turns of the coil. Increasing the number of turns of the coil also increases the resistance of the coil which can negatively affect the voltage sensitivity.
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