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A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by DeltaT in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross - section but of length 2L. The temperature of the wire is raised by the same amount DeltaT in the same time. The volume of N is |
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Answer» Solution :In the first case `((3E)^(2))/(r).t=ms DeltaT"….(1)"` `[H=(V^(2))/(r).t]` When length of the WIRE is doubled, resistance and mass both are doubled THEREFORE, in the second case, `((NE)^(2))/(2R).t=(2m)sDeltaT"……..(2)"` Dividing Eq. (2) by Eq. (1) we get `(N^(2))/(18)=2 or N^(2)=36 or N=6` |
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