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In the circuit the current is to be measured. What is the value of the current if the ammeter shown (a) is a galvanometer with a resistance R_G = 60.00 Omega, (b) is a galvanometer described in (a) but converted to an ammeter by a shunt resistance r_s= 0.02 Omega , (c ) is an ideal ammeter with zero resistance? |
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Answer» Solution :(a) Total resistance in the CIRCUIT is `R_G + 3 = 63 Omega`. HENCE `I = 3/63 = 0.048 A` (B) Resistance of the galvanometer converted to an ammeter is `(R_G r_s)/(R_G + r_s) = (60 Omega xx 0.02 Omega)/((60+ 0.02) Omega) ~~ 0.02 Omega` Total resistance in the circuit is `0.02 Omega + 3 Omega = 3.02 Omega. ` Hence I = 3/3.02 = 0.99 A (c ) For the ideal ammeter with ZERO resistance . I = 3/3 = 1.00 A |
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