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The resistance of an ammeter is 13 Ω and its scale is graduated for a current upto 100A. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto 750A by this meter. The value of shunt resistance is
- a)20 Ω
- b)2 Ω
- c)0.2 Ω
- d)2k Ω
Correct answer is option 'B'. Can you explain this answer?
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The resistance of an ammeter is 13 and its scale is graduated for a c...
Given: Resistance of ammeter, R1 = 13 Ω
Current range of ammeter = 100 A
New current range = 750 A
To increase the current range of the ammeter, a shunt resistor is connected in parallel with the ammeter.
Let the value of the shunt resistance be R2.
Using the formula for equivalent resistance of two resistors in parallel, we get:
1/R = 1/R1 + 1/R2
where R is the equivalent resistance of the ammeter and the shunt resistor.
At the maximum current of 100 A, the entire current flows through the ammeter, so we can write:
R = R1 = 13 Ω
At the maximum current of 750 A, the current through the ammeter is negligible compared to the current through the shunt resistor. Therefore, we can write:
R = R2
Substituting these values in the above equation, we get:
1/R2 = 1/R1 + 1/R2
1/R2 - 1/R2 = 1/R1
R2 = R1/(1 - R1/R)
R2 = 13/(1 - 13/R)
R2 = 13/(R/13 - 1)
R2 = 13(13/R - 1)/R
R2 = 169/R - 13
Substituting R = 2 Ω (from option B), we get:
R2 = 169/2 - 13
R2 = 73 Ω
Therefore, the value of the shunt resistance is 2 Ω (option B).
Current range of ammeter = 100 A
New current range = 750 A
To increase the current range of the ammeter, a shunt resistor is connected in parallel with the ammeter.
Let the value of the shunt resistance be R2.
Using the formula for equivalent resistance of two resistors in parallel, we get:
1/R = 1/R1 + 1/R2
where R is the equivalent resistance of the ammeter and the shunt resistor.
At the maximum current of 100 A, the entire current flows through the ammeter, so we can write:
R = R1 = 13 Ω
At the maximum current of 750 A, the current through the ammeter is negligible compared to the current through the shunt resistor. Therefore, we can write:
R = R2
Substituting these values in the above equation, we get:
1/R2 = 1/R1 + 1/R2
1/R2 - 1/R2 = 1/R1
R2 = R1/(1 - R1/R)
R2 = 13/(1 - 13/R)
R2 = 13/(R/13 - 1)
R2 = 13(13/R - 1)/R
R2 = 169/R - 13
Substituting R = 2 Ω (from option B), we get:
R2 = 169/2 - 13
R2 = 73 Ω
Therefore, the value of the shunt resistance is 2 Ω (option B).
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The resistance of an ammeter is 13 and its scale is graduated for a current upto 100A. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto 750A by this meter. The value of shunt resistance isa)20 b)2 c)0.2 d)2k Correct answer is option 'B'. Can you explain this answer?
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