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In the circuit (Figure ) , the current it to be measured. What is the value of the current if the ammeter shown (a) is galvanometer with a resitance R_(G) = 60,00 Omega, (b) is a galvanometer by a shunt resitance r_(s) = 0.02 omega (c) is an ideal ammter with zero resitance ? |
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Answer» Solution :Total resistance in the CIRCUIT is `R_(G) + 3 = 63 omega` Hence `I = (3)/(63)` = 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 ammerter with ZERO resistance `I = (3)/(3) = 1.00 A` |
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