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A 50 Omegaresistance is connected to a battery of 5 V. A galvanometer of resistance 100 Omega is to be used as an ammeter to measure current through the resistance, for this a resistancer r_(s)is connected to the galvanometer. Which of the following connections should be employed if the measured current is within 1% of the current without the ammeter in the circuit? |
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Answer» `r_(s) = 0.5 Omega` in parallel with the galvanometer `I = V/R = 5/50 = 0.1 A`. When ammeter is added in the circuit then current of the circuit decreases and it is given that current should not CHANGE by more than 1%. Hence current in the circuit when ammeter is included can be written as `I. = 0.099 A` Small variation in current is possible only when effective RESISTANCE of the galvanometer is minimum, So we should CONNECT resistance S in parallel to the galvanometer. Effective resistance of the circuit can be written as follows: `R_("eff") = 50 + (100S)/(100 + S) = V/(I.)` `rArr(100S)/(100 + S) = 5/(0.099) - 50 = 50.50 - 50` `rArr(100S)/(100 + S) = 5/(0.099) - 50 = 50.50 - 50 = 0.5` `rArr 100S - 0.5 (100 + S)` ` rArr100S = 50 + 0.5S` `rArrS = 0.5 Omega` Hence `0.5 Omega` resistance should be connected in parallel to the coil of galvanometer. Hence OPTION (a) is correct. |
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