R=R0(I+K*∂T)
V1/I1<V2/I2 <FROM THE GRAPH)
R1<R2
R0(I+K*∂T1)<R0(I+K*∂T2)
WHERE ∂Ti=Ti-T0
THEREFORE FROM THE ABOVE RELATION
T1<T2
graphical analysis
M1<M2
WHERE M IS THE SLOPE
Mi=Vi/Ii
The I-V characteristics of a metal wire at two temperatures is shown, then
a) T1>T2
b) T2>T1
c) T1=T2
d) The relation may vary
The slope here is giving the magnitude of conductance, i think. so Resistance at T1 is less than resistance at T2. so (b ).
R=R0(I+K*∂T)
V1/I1<V2/I2 <FROM THE GRAPH)
R1<R2
R0(I+K*∂T1)<R0(I+K*∂T2)
WHERE ∂Ti=Ti-T0
THEREFORE FROM THE ABOVE RELATION
T1<T2
graphical analysis
M1<M2
WHERE M IS THE SLOPE
Mi=Vi/Ii