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A galvanometer having 30 divisions has current sensitivity of 20 muA/ division. It has a resistance of 25 ohm. How will you convert it into an ammeter measuring voltmeter reading upto 1V? |
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Answer» Solution :The full scale DEFLECTION current `i_g = 30 XX (20 xx 10^(-6) ) = 6 xx 10^(-4) A` If S is the required VALUE of the shunt connected in parallel with galvanometer, then ` i_g = (S)/(S+G) I rArr 6 xx 10^(-4) = (S)/(S +25) xx 1` After solving , we get `S= 150/9994 Omega = 0.0150 Omega` The resistance of the ammeter ` 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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