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A galvanometer having 30 divisions has current sensitivity of 20 mu A // division. It has a resistance of 25 ohm. How will you convert it into an ammeter measuring upto 1 A ? How wil you nowconvert this ammeter into a voltmeter reading upto 1 V ? |
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Answer» Solution :The full scale deflection current `i_g = 30 XX(20 xx 10^(-6))` `= 6 xx 10^(-4)A.` It S is the required value of the SHUNT connected in parallel with the galvanometer, then `i_g = (S)/(S+G) i or 6 xx 10^(-4) = (S)/(S +25) xx 1` After solving , we get `S = (150)/(9994) Omega = 0.0150 Omega` The resistance of the ammter `R_A = (SG)/(S+G) = (0.0150 xx 25)/(0.0150+ 25) = 0.0150 Omega` To convert this ammeter into the voltmeter , we can use `V = i_(g) (R_A + R_0)` Here `V = 1V, i_g = 1A` `THEREFORE 1 = 1(0.0150 + R_0) or R_0 = 0.985 Omega` |
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