%3Cp%3E%3Cspan%20class%3D%22c2%22%3E%3Cstrong%3E%3Cspan%20class%3D%22c1%22%3EPLE
ID: 3898878 • Letter: #
Question
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/ Answer
resistance = 30 ohm
So current = 166.67 milli ampere....Answer a
resistance = 9.26 ohm
so current = 540 milli ampere....Answer b
resistance = 3.33 ohm
so current = 1500 milli ampere .... Answer c
constant slope.....since temperature is fixed, so is resistance
The current that would flow through the block in each case is SMALLER THAN you found in part a, by a factor of (1 + aT) = 1.56 times
V/I graph slope constant , steeper than in part (a)....because resistance has increased
if brought from room T to 160 oC solely via internal electrical heating, then V/I graph increasing slope....becauseresistance increasing with temperature
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I*(rho) = constant ,so new rho (i.e resistivity) = 1.44 Ohm-metre .....Answer
I*(1+aT) = constant , so new a = 0.0249 Kelvin inverse......Answer
if this second block is brought from room T to 160 oC solely via internal electrical heating, then also the V-I graph will have INCREASING SLOPE
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