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A galvanometer of resistance G is converted into a voltmenter to measrue upto V volts by connecting a resistance R_(1) in series with the coil. If a resistance R_(2) is connected series with it, then it can measure upto 1//2 volts. Find the resistance, in terms of R_(1) and R_(2), required to be connected to convert in into a voltmeter that can read upto 2V. Also find the resistance G of the galvanometer in terms of R_(1) and R_(2) |
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Answer» Solution :Let G be the resistance of galvanometer and `l_(g)` be the CURRENT through galvanometer for full scale DEFLECTION. If this galvanometer is converted into voltmenter to measure upto V volts, then `R_(1)=(Vl2)/(l_(g))-G=(V)/(2l_(g))-G` Substracting (ii) from (i), we get `R_(1)-R_(2)=(V)/(l_(g))-(V)/(2l_(g))=(V)/(2l_(g))` or `(V)/(l_(g))=2(R_(1)-R_(2))` Multiply (ii) with 2 and substracting from (i), we get `R_(1)-2R_(2)=-G+2G=G, i.e., G=R_(1)-2R_(2)` Let `R_(3)` resistance be connected in series to CONVERT the galvanometer into a voltmenter of range 0 to 2V then `R_(3)=(2V)/(l_(g))-G=2xx2(R_(1)-R_(2))-(R_(1)-2R_(2))=3R_(1)-2R_(2)` |
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