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A moving coil galvanometer is converted into an ammeter reading upto 0.03 A by connecting a shunt of resistance 4r across it and into ammeter upto 0.06 A, when a shunt of resistance r is used, what is the maximum current which can be sent through this galvanometer if no shunt is used
- a)0.04A
- b)0.03A
- c)0.02A
- d)0.01A
Correct answer is option 'C'. Can you explain this answer?
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A moving coil galvanometer is converted into an ammeterreading upto 0....
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A moving coil galvanometer is converted into an ammeterreading upto 0....
Given:
Shunt resistance for 0.03 A reading, S1 = 4r
Shunt resistance for 0.06 A reading, S2 = r
To find: Maximum current that can be sent through the galvanometer without using a shunt
Concept Used:
Shunt resistance (S) is given by the formula:
S = (G-1)R
Where G is the sensitivity of the galvanometer and R is the resistance of the galvanometer.
The formula for the maximum current that can be sent through the galvanometer without using a shunt is:
Imax = G/2R
Solution:
Let the resistance of the galvanometer be Rg and the sensitivity be G.
For 0.03 A reading:
Total current flowing through the circuit, I = 0.03 A + I1
where I1 is the current flowing through the shunt resistance S1.
By applying Kirchhoff's current law:
I = I1 + I2
where I2 is the current flowing through the galvanometer.
I2 = I - I1
I2 = 0.03 A + I1 - I1
I2 = 0.03 A
The shunt resistance S1 is given by:
S1 = (G/Rg - 1)R
4r = (G/Rg - 1)R
G/Rg = 1 + 4r/R
For 0.06 A reading:
Total current flowing through the circuit, I = 0.06 A + I2
where I2 is the current flowing through the shunt resistance S2.
By applying Kirchhoff's current law:
I = I1 + I2
I1 = I - I2
I1 = 0.06 A - I2
The shunt resistance S2 is given by:
S2 = (G/Rg - 1)R
r = (G/Rg - 1)R
G/Rg = 1 + r/R
Dividing the two equations, we get:
(1 + 4r/R)/(1 + r/R) = 2
R = 3r
The maximum current that can be sent through the galvanometer without using a shunt is given by:
Imax = G/2R
Imax = G/2(3r)
Imax = G/6r
Substituting the value of G/Rg from the equation above, we get:
Imax = (1 + 4r/R)/6r
Imax = (1 + 4r/3r)/6r
Imax = 0.02 A
Therefore, the maximum current that can be sent through the galvanometer without using a shunt is 0.02 A, which is option (c).
Shunt resistance for 0.03 A reading, S1 = 4r
Shunt resistance for 0.06 A reading, S2 = r
To find: Maximum current that can be sent through the galvanometer without using a shunt
Concept Used:
Shunt resistance (S) is given by the formula:
S = (G-1)R
Where G is the sensitivity of the galvanometer and R is the resistance of the galvanometer.
The formula for the maximum current that can be sent through the galvanometer without using a shunt is:
Imax = G/2R
Solution:
Let the resistance of the galvanometer be Rg and the sensitivity be G.
For 0.03 A reading:
Total current flowing through the circuit, I = 0.03 A + I1
where I1 is the current flowing through the shunt resistance S1.
By applying Kirchhoff's current law:
I = I1 + I2
where I2 is the current flowing through the galvanometer.
I2 = I - I1
I2 = 0.03 A + I1 - I1
I2 = 0.03 A
The shunt resistance S1 is given by:
S1 = (G/Rg - 1)R
4r = (G/Rg - 1)R
G/Rg = 1 + 4r/R
For 0.06 A reading:
Total current flowing through the circuit, I = 0.06 A + I2
where I2 is the current flowing through the shunt resistance S2.
By applying Kirchhoff's current law:
I = I1 + I2
I1 = I - I2
I1 = 0.06 A - I2
The shunt resistance S2 is given by:
S2 = (G/Rg - 1)R
r = (G/Rg - 1)R
G/Rg = 1 + r/R
Dividing the two equations, we get:
(1 + 4r/R)/(1 + r/R) = 2
R = 3r
The maximum current that can be sent through the galvanometer without using a shunt is given by:
Imax = G/2R
Imax = G/2(3r)
Imax = G/6r
Substituting the value of G/Rg from the equation above, we get:
Imax = (1 + 4r/R)/6r
Imax = (1 + 4r/3r)/6r
Imax = 0.02 A
Therefore, the maximum current that can be sent through the galvanometer without using a shunt is 0.02 A, which is option (c).
Community Answer
A moving coil galvanometer is converted into an ammeterreading upto 0....
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A moving coil galvanometer is converted into an ammeterreading upto 0.03 A by connecting a shunt ofresistance 4r across it and into ammeter upto 0.06 A,when a shunt of resistance r is used, what is the maximumcurrent which can be sent through this galvanometerif no shunt is useda)0.04Ab)0.03Ac)0.02Ad)0.01ACorrect answer is option 'C'. Can you explain this answer?
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