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A galvanometer of resistance `50 Omega ` is connected to a battery of `3 V` along with resistance of `2950 Omega` in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 division the above series resistance should beA. `4450 Omega`B. `5050 Omega`C. `5550 Omega`D. `6050Omega` |
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Answer» Correct Answer - A Current decreases `20/30` times or `2/3` times. Therefore, net resistance should become `3/2` times. `:. R+50=3/2(2950+50)` Solving we get, `R=4450Omega` |
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