Question 3.5: a) A 50 mV, 1 mA d’ Arsonval movement is to be used in an am...

a) A 50 mV, 1 mA d’Arsonval movement is to be used in an ammeter with a full-scale reading of 10 mA. Determine R_{A}.
b) Repeat (a) for a full-scale reading of 1A.
c) How much resistance is added to the circuit when the 10 mA ammeter is inserted to measure current?
d) Repeat (c)for the 1A ammeter.

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

a) From the statement of the problem, we know that when the current at the terminals of the ammeter is 10 mA, 1 mA is flowing through the meter coil, which means that 9 mA must be diverted through R_{A} We also know that when the movement carries 1 mA, the drop across its terminals is 50 mV. Ohm’s law requires that

90\times 10^{-3}R_{A}=50\times 10^{-3}

or

R_{A}=\frac{50}{9}=5.555\Omega .

b) When the full-scale deflection of the ammeter is 1 A, R_{A} must carry 999 mA when the movement carries 1 mA. In this case, then,

999\times 10^{-3}R_{A}=50\times 10^{-3},

or

R_{A}=\frac{50}{999} \approx 50.05m\Omega.

c) Let R_{m} represent the equivalent resistance of the ammeter. For the 10 mA ammeter,

R_{m}=\frac{50 mV}{10 mA}=5\Omega ,

or, alternatively,

R_{m}=\frac{\left(50\right)\left(\frac{50}{9} \right) }{50+\left(\frac{50}{9} \right)}=5\Omega ,

d) For the 1 A ammeter

R_{m}=\frac{50 mV}{1 A} =0.050\Omega ,

or, alternatively,

R_{m}=\frac{\left(50\right)\left(\frac{50}{999} \right) } {50+\left(\frac{50}{999} \right)}=0.050\Omega ,

 

 

 

Related Answered Questions