Pages

Sunday, September 8, 2013

An ammeter has a current range of 0 - 5 A, and its internal resistance is 0.2Ω . In order to change the range to 0 - 25 A, we need to add a resistance of

An ammeter has a current range of 0 - 5 A, and its internal resistance is 0.2Ω . In
order to change the range to 0 - 25 A, we need to add a resistance of  
                 (GATE-EE-2010)

(A) 0.8Ω in series with the meter

(B) 1.0Ω in series with the meter
(C) 0.04Ω in parallel with the meter 
(D) 0.05Ω in parallel with the meter



Ans: D



Explanatory answer :


In ammeter designs, external resistors added to extend the usable range of the movement are connected in parallel with the movement rather than in series as is the case for voltmeters. This is because we want to divide the measured current, not the measured voltage, going to the movement, and because current divider circuits are always formed by parallel resistances.

The shunt resistance can be calculated as follows

Let  Rm = internal resistance of coil=0.2 Ω   
 Rsh  = low shunt resistance going to add                               


Im = full scale deflection current=5A
Ish = shunt current    

As the two resistances are in parallel, the voltage drop across them is same

therefore    IshRsh  = IRm 

 Rsh  =  IRm /Ish 

Rsh   = 5* 0.2 / (25-5)

Rsh  = 0.05  Ω    is connected in parallel with the meter


2 comments: