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A galvanometer with `50` divisions on the scale has a resistance of `25 Omega`. A current of `2 xx 10^-4` A gives a deflection of one scale division. The additional series required to convert it into a voltmeter reading up to `25 V` is.A. `1200 Omega`B. `1225 Omega`C. `2475 Omega`D. `2500 Omega` |
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Answer» Correct Answer - C `i_g = 50 xx 2 xx 10^-4 = 10^-2 A, G = 25 Omega` `V = i_g(R + G)` `25 = 10^-2(R + 25) rArr R = 2475 Omega`. |
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