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A wire of 15 Omega resistance is gradually stretched to double its original length. It is then cut into two equal parts. These parts are then connected in parallel across a 3.0 volt battery. Find the current drawn from the battery. |
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Answer» Solution :Here R = 15 `Omega`. When the wire is gradually stretched to double its original length, its cross-sectionarea is reduced to half of its original value so that the volume remains same. Hence l = 21 and A. = A/2. ` therefore ` New resistance of wire`R. = (rho l.)/(A.) = (rho . (2l))/((A/2)) = 4 (rho l)/(A) = 4R = 4 XX 15 = 60 Omega` On cutting the wire into 2 equal parts, resistance of each part `R_1 = R_2= (R.)/(2) = 30 Omega` On joining the two parts in PARALLEL, equivalent resistance `R_(eq) = (R_1 R_2)/(R_1 + R_2) = (30 xx 30)/(30 + 30) = 15 Omega` If emf of battery be `epsi`= 3.0 V, then current DRAWN from the battery `I = (epsi)/(R_(eq)) = (3.0 V)/(15Omega) = 0.2 A` |
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