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A galvanometer of resistance 50Omega is connecter to a battery of 8 V along with a resistance of 3950Omega in series. A full scale deflection on 30 div is obtained in the galvanometer. In order to reduce this deflection of 15 division, the resistance in series should be ______ Omega. |
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Answer» 7900 `I=8/(3950+50)=8/4000=1/500A` Now, `I/2=V/(R^(1)+G)` `1/(2xx500)=8/(R^(1)+50)` `thereforeR^(1)+50=8xx1000` `thereforeR^(1)=8000-50` `thereforeR^(1)=7950Omega` |
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