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An ammeter of resistance 0.80Ω can measure currents upto 1 A . (i) What must be the shunt resistance to enable the ammeter to measure current upto 5A ?

(ii) What is the combined resistance of the ammeter and the shunt?



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We have to make the current flowing in the internal resistance 0.80 Ω to 1/5 th of the current flowing ie, of 5 A.

When 1 A flows through the internal resistance 0.80 Ω, the remaining 4 Amp current flows through the shunt resistance S.   Their currents are inversely proportional to their resistances, (they are in parallel)

    so      0.80 Ω/ S  =  4 A /1 A  
               S = 0.2 Ω  = shunt resistance
Combined resistance of shunt and internal resistance:  0.8 * 0.2 / (0.8+0.2) = 0.16 Ohms

1 5 1
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