1.

In the circuit shown in figure, the current is to measured. What is the value of the current if the ammeter shown (i) is a galvanometer with a resistance `R_G=60*00Omega`. (ii) is a galvanometer described in (i) but converted to an ammeter by a shunt resistance `r_s=0*02Omega`, (iii) is an ideal ammeter with zero resistance?

Answer» Let G be the resistance of ammeter (i.e. galvanometer). Then current in the circuit,
`I=(epsilon)/(R+G)=(3*00)/(3*00+60*00)=0*048A`.
(ii) When the ammeter (i.e., galvanometer) is shunted with resistance S, its effective resistance,
`R_p=(GS)/(G+S)=(60xx0*02)/(60+0*02)~~0*02Omega`
Current in the circuit,
`I=(epsilon)/(R+R_p)=(3*00)/(3*00+*02)~~0*99A`.
(iii) For the ideal ammeter with zero resistance current, `I=(3*00)/(3*00)=1*00A`.


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