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A galvanometer having 30 divisions has current sensitivity of 20 μA/ division. It has a resistance of 25 ohm. How will you convert it into an ammeter measuring upto 1 A? How will you now convert this ammeter into a voltmeter reading upto 1 V? |
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Answer» Solution :The full scale deflection current `i_(g)=30xx(20xx10^(-6))=6xx10^(-4)A`. If S is the required VALUE of the shunt connected in parallel with the galvanometer, then `i_(g)=S/(S+G)irArr6xx10^(-4)=S/(S+25)xxl` After SOLVING, we GET `S=150/9994Omega-0.01150Omega` The resistance of the ammeter `R_(A)=(SG)/(S+G)=(0.0150xx25)/(0.0150+25)=0.0150Omega` To convert this ammeter into the voltmeter, we can use `V=i_(g)(R_(A)+R_(0))` Here `V=1V,i_(g)=1A` `therefore1=1(0.0150+R_(0))` or `R_(0)=0.985Omega` |
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