Answer:
a)
[tex] \bf \frac{\partial R}{\partial R_1}=\frac{R_2^2}{(R_1+R_2)^2}[/tex]
[tex] \bf \frac{\partial R}{\partial R_2}=\frac{R_1^2}{(R_1+R_2)^2}[/tex]
b) and c)
See explanation below
Step-by-step explanation:
a)
The combined resistance R is a function of the the two
parallel resistances
[tex] \bf \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{R_1+R_2}{R_1R_2}[/tex]
So
[tex] \bf R(R_1,R_2)=\frac{R_1R_2}{R_1+R_2}[/tex]
and we have
[tex] \bf \frac{\partial R}{\partial R_1}=\frac{R_2(R_1+R_2)-R_1R_2}{(R_1+R_2)^2}=\frac{R_2^2}{(R_1+R_2)^2}[/tex]
Similarly,
[tex] \bf \frac{\partial R}{\partial R_2}=\frac{R_1(R_1+R_2)-R_1R_2}{(R_1+R_2)^2}=\frac{R_1^2}{(R_1+R_2)^2}[/tex]
b)
If we divide both the numerator and denominator by [tex] \bf R_1[/tex] in the expression for R, we get
[tex] \bf R=\frac{R_2}{1+R_2/R_1}[/tex]
hence, if we held [tex] \bf R_2[/tex] constant and increase [tex] \bf R_1[/tex] the fraction [tex] \bf \frac{R_2}{R_1}[/tex] gets smaller and so does the denominator of R, as a consequence R gets larger.
When [tex] \bf R_1[/tex] is very large, the denominator of R is close to 1, so R is close to [tex] \bf R_2[/tex]
c)
By a symmetric reasoning, we see that R gets larger when holding [tex] \bf R_1[/tex] constant and [tex] \bf R_2[/tex] increases.
In this case, R gets closer to [tex] \bf R_1[/tex] as [tex] \bf R_2[/tex] grows.
