Answer to Question #92775 in Electric Circuits for Sridhar

Question

If only 1/51 th of the main current is to be passed through a galvanometer then the shunt required is R1 and if only 1/11 th of the main voltage is to be developed across the galvanometer, then the resistance required is R2. Then R2/R1=

Answer 1

$V=V_G=V_{R1}, \space I=I_G+I_{R1}$

$I_G=\frac {1}{51} I, \space I_{R1}=\frac {50}{51}I$

$V_G=I_G R_G=V_{R1}=I_{R1}R1$

$R_G=\frac{I_{R1}}{I_G}R1=50R1 \qquad (1)$

$V=V_G+V_{R2}, \space I=I_G=I_{R2}$

$V_G=\frac {1}{11}V, \space V_{R2}=\frac {10}{11}V$

$I_G=\frac{V_G}{R_G}=I_{R2}=\frac{V_{R2}}{R2}$

$R_G=\frac{V_G}{V_{R2}}R2=\frac{1}{10}R2 \qquad (2)$

from (1) and (2)

$50R1=\frac{1}{10}R2 \space \rightarrow \space \frac{R2}{R1}=500$

Answer: R2/R1=500