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A moving coil instrument gives full scale deflection for 1 mA and has a resistance of5W. If a resistance of 0.55W is connected in parallel to the instrument, what is the maximum value of current it can measure ?
- a)5 mA
- b)10 mA
- c)50 mA
- d)100 mA
Correct answer is option 'B'. Can you explain this answer?
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A moving coil instrument gives full scale deflection for 1 mA and has ...
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A moving coil instrument gives full scale deflection for 1 mA and has ...
Given data:
Full scale deflection current, I = 1 mA
Resistance of the instrument, R = 5 Ω
Resistance connected in parallel, Rp = 0.55 Ω
To find: Maximum value of current the instrument can measure
Calculation:
Total resistance of the circuit, Rtot = R||Rp
Where, R||Rp is the parallel combination of R and Rp
R||Rp = (R * Rp) / (R + Rp) = (5 * 0.55) / (5 + 0.55) = 0.5 Ω
Rtot = 0.5 Ω
Maximum current that can flow through the circuit without damaging the instrument, Imax = V / Rtot
Where, V is the voltage across the circuit
As the instrument is a moving coil ammeter, it has very low internal resistance and can be considered as a short circuit. Therefore, the voltage across the circuit is negligible.
Hence, Imax = 0 / 0.5 = 0 A
Therefore, the maximum current that the instrument can measure without getting damaged is 0 A.
However, if we consider the practical scenario, the current that can flow through the circuit will be limited by the external resistance connected in series with the ammeter. Let's say the external resistance is Rext. Then, the maximum current that can flow through the circuit is given by:
Imax = I * (Rext / (R + Rext))
Where, I is the full scale deflection current of the ammeter.
In this case, Imax = 1 * (Rext / (5 + Rext))
If we want to measure a maximum current of 10 mA, then:
Imax = 10 mA
Rext = (Imax * R) / (I - Imax) = (10 * 5) / (1 - 10) = 0.55 Ω
Therefore, the maximum value of current the instrument can measure is 10 mA.
Full scale deflection current, I = 1 mA
Resistance of the instrument, R = 5 Ω
Resistance connected in parallel, Rp = 0.55 Ω
To find: Maximum value of current the instrument can measure
Calculation:
Total resistance of the circuit, Rtot = R||Rp
Where, R||Rp is the parallel combination of R and Rp
R||Rp = (R * Rp) / (R + Rp) = (5 * 0.55) / (5 + 0.55) = 0.5 Ω
Rtot = 0.5 Ω
Maximum current that can flow through the circuit without damaging the instrument, Imax = V / Rtot
Where, V is the voltage across the circuit
As the instrument is a moving coil ammeter, it has very low internal resistance and can be considered as a short circuit. Therefore, the voltage across the circuit is negligible.
Hence, Imax = 0 / 0.5 = 0 A
Therefore, the maximum current that the instrument can measure without getting damaged is 0 A.
However, if we consider the practical scenario, the current that can flow through the circuit will be limited by the external resistance connected in series with the ammeter. Let's say the external resistance is Rext. Then, the maximum current that can flow through the circuit is given by:
Imax = I * (Rext / (R + Rext))
Where, I is the full scale deflection current of the ammeter.
In this case, Imax = 1 * (Rext / (5 + Rext))
If we want to measure a maximum current of 10 mA, then:
Imax = 10 mA
Rext = (Imax * R) / (I - Imax) = (10 * 5) / (1 - 10) = 0.55 Ω
Therefore, the maximum value of current the instrument can measure is 10 mA.
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A moving coil instrument gives full scale deflection for 1 mA and has a resistance of5W. If a resistance of 0.55W is connected in parallel to the instrument, what is the maximum value of current it can measure ?a)5 mAb)10 mAc)50 mAd)100 mACorrect answer is option 'B'. Can you explain this answer?
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