When two electrical resistors with resistanceR1>0 andR2>0 are wired in parallelin a circuit, the combined resistanceR, measured in ohms (Ω), is given by1R=1R1+1R2(a) Find∂R∂R1and∂R∂R2after solving forR(e.g.,R=. . .).(b) Describe how an increase inR1withR2held constant affectsR. (WillRincreaseor decrease?)(c) Describe how a decrease inR2withR1held constant affectsR. (WillRincreaseor decrease?)

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 belowStep-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.