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A galvanometer of resistance 20 Omega is shunted by a 2 Omega resistor. What part of the main current flows through the galvanometer ? |
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Answer» Solution :`i_g/i = (S)/(G +S)`. Given `G = 20 Omega , S = 2 Omega` `therefore i_g/i = 2/22 = 1/11 , 1/11` th PART of CURRENT passes through galvanometer. |
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