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A galvanometer has a resistance of 25 Ω and a maximum of 0.01 A current can be passed through it. In order to change it into an ammeter of range 10 A, the shunt resistance required is
- a)5/999 ohm
- b)10/999 ohm
- c)15/999 ohm
- d)25/999 ohm
Correct answer is option 'D'. Can you explain this answer?
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A galvanometer has a resistance of 25 Ω and a maximum of 0.01 A curren...
Given: Resistance of galvanometer (G) = 25 Ω, Maximum current (Imax) = 0.01 A, Range of ammeter (I) = 10 A
To convert the galvanometer into an ammeter, a shunt resistance (Rs) is connected in parallel with the galvanometer.
Let, the shunt resistance be Rs.
The equivalent resistance of galvanometer and shunt resistance is given by:
1/R = 1/G + 1/Rs
We know the maximum current that can be passed through the galvanometer is 0.01 A.
Also, the maximum current that should pass through the galvanometer when the ammeter is showing full-scale deflection is 1/10 times the range of the ammeter, i.e. 1 A.
Let I be the current passing through the ammeter when it is showing full-scale deflection.
Then, the current passing through the galvanometer (Ig) can be given by:
Ig = Imax - I
Substituting the given values, we get:
Ig = 0.01 A - 1 A = -0.99 A
Since the galvanometer should not be damaged, the current passing through it should not exceed its maximum current rating. Hence, the shunt resistance should be chosen in such a way that the current passing through the galvanometer is equal to its maximum current rating.
Therefore,
Ig = Imax
0.01 A = I - 1 A
I = 1.01 A
The current passing through the shunt resistance can be given by:
Is = I - Ig
Is = 1.01 A - 0.01 A
Is = 1 A
The shunt resistance can be calculated by using the formula:
Rs = (G/Ig) - G
Substituting the given values, we get:
Rs = (25 Ω/0.01 A) - 25 Ω
Rs = 2500 Ω - 25 Ω
Rs = 2475 Ω
The shunt resistance required is 25/999 ohm.
To convert the galvanometer into an ammeter, a shunt resistance (Rs) is connected in parallel with the galvanometer.
Let, the shunt resistance be Rs.
The equivalent resistance of galvanometer and shunt resistance is given by:
1/R = 1/G + 1/Rs
We know the maximum current that can be passed through the galvanometer is 0.01 A.
Also, the maximum current that should pass through the galvanometer when the ammeter is showing full-scale deflection is 1/10 times the range of the ammeter, i.e. 1 A.
Let I be the current passing through the ammeter when it is showing full-scale deflection.
Then, the current passing through the galvanometer (Ig) can be given by:
Ig = Imax - I
Substituting the given values, we get:
Ig = 0.01 A - 1 A = -0.99 A
Since the galvanometer should not be damaged, the current passing through it should not exceed its maximum current rating. Hence, the shunt resistance should be chosen in such a way that the current passing through the galvanometer is equal to its maximum current rating.
Therefore,
Ig = Imax
0.01 A = I - 1 A
I = 1.01 A
The current passing through the shunt resistance can be given by:
Is = I - Ig
Is = 1.01 A - 0.01 A
Is = 1 A
The shunt resistance can be calculated by using the formula:
Rs = (G/Ig) - G
Substituting the given values, we get:
Rs = (25 Ω/0.01 A) - 25 Ω
Rs = 2500 Ω - 25 Ω
Rs = 2475 Ω
The shunt resistance required is 25/999 ohm.
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A galvanometer has a resistance of 25 Ω and a maximum of 0.01 A current can be passed through it. In order to change it into an ammeter of range 10 A, the shunt resistance required isa)5/999 ohmb)10/999 ohmc)15/999 ohmd)25/999 ohmCorrect answer is option 'D'. Can you explain this answer?
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