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(a) State the principle of working of a galvanometer. (b) Agalvanometer of resistance G is converted into avoltmeter to measure up to V volts by connecting a resistance R_(1) , in series with the coil. If a resistance R_(2)is connected in series with it, then it can measure up to V/2 volts. Find the resistance, in terms of R_(1)"and R_(2), required to be connected to convert it into a voltmeter that can read up to 2 V. Also find the resistance G of the galvanometer in terms of R_(1)" and " R_(2). |
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Answer» Solution :(b) Ist case `I_(g)` = MAXIMUM CURRENT through galvanometer `R_(1) = V/I_(g) - G `….(i) `2^(nd) "case "` `R_(2) = V/(2I_(g)) - G `….(ii) `3^(rd) ` case ` R = (2V)/I_(g) - G ` ….(iii) From eq. (i) and (ii) `R_(1) - R_(2) = V/I_(g) - G - V/(2I_(g)) + G = V/(2I_(g))` `2(R_(1) - R_(2)) = V/I_(g)` Putting the value of `V/I_(g)` in (i) we GET `R_(1) = 2(R_(1) - R_(2)) - G` ` R_(1) = 2R_(1) - 2R_(2) - G` ` G = 2R_(1) - 2R_(2) - R_(1)` ` = R_(1) - 2R_(2)` Putting the value of `V/I_(g)` and G in eq. (iii) we get : `R = 2[2(R_(1) - R_(2))] - (R_(1) - 2R_(2))` ` = 4R_(1) - 4R_(2) - R_(1) + 2R_(2)` ` = 3R_(1) - 2R_(2)` |
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