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When 50Omega resistance is connected to one voltmeter, its range is obtained upto V volt. Now, instead of 50Omega, when 500Omega resistance is connected to this voltmeter, its range is obtained upto 2 V volt. Find resistance of this voltmeter. |
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Answer» `100Omega` (i) `R_(1)=G(n_(1)-1)` `therefore50=G(n_(1)-1)=Gn_(1)-G""...(2)` (ii) `R_(2)=G(n_(2)-1)` `therefore500=G(2n_(1)-1)=2Gn_(1)-G""...(3)` Multiplying equation (2) by 2, we get, `100=2Gn_(1)-2G""...(4)` Subtracting equation (4) from equation (3) we get, 400 = G `thereforeG=400Omega` |
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