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InterviewSolution
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Two wires of different materials P and Q have resistance per unit lengths `50 Omega km^(-1)` and `25 Omega km^(-1)` and temperature coefficients of resistance, `0.0025 ^@C^(-1)` and `0.00075^@C^(-1)` respectively. If it is desired to make a coil having `700 Omega` resistance and a temperature coefficient of resistance `0.001^@C^(-1)` by using suitable length of two wires in series. Calculate their respective length. |
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Answer» Let `R_(P) and R_(Q)` be the resistance of the two wires at the given temperature and `l_(P) and l_(Q)` be their respective lengths. If `Delta T` be the change in temperature, then the total resistance of two wires in series is `R = R_(P) ( 1 + 0.00025 Delta T) + R_(Q) (1 + 0.00075 Delta T)` As `0.0001^@C^(-1)` is the temperature coefficient of resistance of the combination of two wires in series, so `R = (R_(P) + R_(Q)) [1 + 0.001 Delta T]` From (i) and (ii), we have `= R_(P) ( 1 +0.0025 Delta T) +R_(Q) (1 +0.00075 Delta T) = (R_(P) + R_(Q)) [1 + 0.001 Delta T]` or `R_(P) (0.0015 Delta T) = R_(Q) (0.00025 Delta T)` or `R_(Q) = 150/25 R_(P) = 6 R_(P)` Since `R_(P) + R_(Q) = 700` `:. R_(P) + 6 R_(P) = 700 or R_(P) = 100 Omega` and `R_(Q) = 600 Omega` `:.` length of wire `P, l_(P) = 1/50 xx 100 = 2 km` length of wire Q, `l_(Q) = 1/25 xx 600 = 24 km` |
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