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(a) State the principle of working of a galvonometer. (b) A galvonometer of resistance G is converted into a voltmeter to measure upto 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 upto 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 upto 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) As per question a galvanometer of RESISTANCE G is converted into a voltmeter of range V by connecting a series resistance `R_1`, so `V = I_(g) (G + R_1) ""…..(i)` And it is converted into a volmeter of range `V/2` by connecting a series resistance `R_2`, so, `V/2 = I_(g) (G + R_2)"".......(II)` To convert into a voltmeter of range 2 V we should connect a resistance R in series of it. Then, `2V = I_(g)(G + R) ""......(iii)` DIVIDING (i) by (ii), we have `2 = (G + R_1)/(G + R_2) = G = (R_1 - 2R_2)` Again dividing (iii) by (i), we GET `2 = (G + R)/(G + R_1) implies R = (G + 2R_1) = (R_1 - 2R_2) = (R_1 - 2R_2) = (3R_1 - 2R_2)`. |
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