4 problems involving resistance, tungsten filaments, current, emf
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With very detailed computations, the problems are solved.
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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 ...
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- BEng, Allahabad University, India
- MSc , Pune University, India
- PhD (IP), Pune University, India
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