A galvanometer of resistance `50Omega` is connected to a b attery of 3V alongwith a resistance of `2950Omega` in series. A full scale deflection of 30 division is obtained in the galvanometer in order to reduce this deflection to 20 division. The resistance in sereis. should be:-A. `5050Omega`B. `550Omega`C. `6050Omega`D. `4450Omega`

A galvanometer of resistance `50Omega` is connected to a b attery of 3

1.	A galvanometer of resistance `50Omega` is connected to a b attery of 3V alongwith a resistance of `2950Omega` in series. A full scale deflection of 30 division is obtained in the galvanometer in order to reduce this deflection to 20 division. The resistance in sereis. should be:-A. `5050Omega`B. `550Omega`C. `6050Omega`D. `4450Omega`
Answer» Correct Answer - D Current through the galvanomete, `l=(3)/((50+2950))=10^(-3)A` Current for 30 divisions `= 10^(-3)A` Current for 20 divisions `= (10^(-3))/(30)xx20 =(2)/(3)xx10^(-3)A` For the same deflection to obtain for 20 divisions, let resistance added be R `therefore " " (2)/(3)xx10^(-3)=(3)/((50+1R))` `rArr " " R = 4450 Omega`

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