1.

In the circuit, the current is to be measured. What is the value of the current if the ammeter shown : (b) is a galvanometer described in (i) but converted to an ammeter by a shunt resistance r_s = 0.02 Omega

Answer»

Solution :(b) RESISTANCE of the galvanometer as AMMETER is
`(R_(0) r_s)/(R_(G) r_s) = (60Omega XX 0.02 OMEGA)/((60+0.02))= 0.02 Omega`
Total resistance `R = 0.02Omega + 3OMEGA = 3.02Omega`
Hence, `I = (3)/(302) = 0.99 A`


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