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State the principle of working of a galvanometer. A galvanometer of resistance G is converted into a voltmeter to measure upto V volts by connecting resistance R_(1) in series with the coil. If a resistance R_(2) isconnected in series with it, then it can measure V/2 volts. Find the resistance, in terms of R_(1) and R_(2), required to be connected to convert it into a volt meter 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 :Energy stored in a CAPACITOR, `E=(1)/(2)CV^(2)` In parallel,`0.25=(1)/(2)(C_(1)+C_(2))(100)^(2)` In series,`0.045 =(1)/(2)((C_(1)C_(2))/(C_(1)+C_(2)))(100)^(2)` From (i)`C_(1)+C_(2)=0.25xx2xx10^(-4)` `C_(1)+C_(2)=5xx10^(-5)` From (II)`(C_(1)C_(2))/(C_(1)+C_(2))0.045xx2xx10^(-4)` `(C_(1)C_(2))/(C_(1)+C_(2))=0.09xx10^(-4)=9xx10^(-6)` Form (iii) `C_(1)C_(2)=(2xx0.045xx5xx10^(-5))/(10^(4))=4.5xx10^(-10)` `C_(1)-C_(2)=sqrt((C_(1)+C_(2))^(2)-4C_(1)C_(2))` `C_(1)-C_(2)=2.64xx10^(-5)` SOLVING (ii) and (iv)`C_(1)=38.2 mu F` `C_(2)=11.8 mu F` In parallel`Q_(1)=C_(1)V=38.2xx10^(-6)xx100=38.2xx10^(-4)C` `Q_(2)=C_(2)V=11.8xx10^(-6)xx100=11.2xx10^(-4)C` |
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