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    4 problems involving resistance, tungsten filaments, current, emf

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    1.) because thermal coeff. of electrical resistivity of tungsten (a):
    a = 0.0046 /degree centigrade

    at 20 degree centigrade:
    V = i*R
    V = 100 volt
    i = 6.0 amp.
    R = (3 + ri) ohm
    where ri is the internal resistance of battery.
    => 100 = 6 * ( 3 + ri)
    => ri = (50/3) - 3 = 13.67 ohm
    Now at 2000 deg C:

    R2000 = R20 * {1 + a * (2000 -20)}
    = 3 * (1 + 0.0046 * 1980) = 30.32 ohm

    therefore at 2000 deg C:
    i = V /( ri + R2000) = 100/(13.67 + 30.32) = 2.27 A Ans

    P = i^2 *R = (2.27)^2 ...

    Solution Summary

    With very detailed computations, the problems are solved.

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