Question

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 1

Answer:

a)

$\bf \frac{\partial R}{\partial R_1}=\frac{R_2^2}{(R_1+R_2)^2}$

$\bf \frac{\partial R}{\partial R_2}=\frac{R_1^2}{(R_1+R_2)^2}$

b) and c)

See explanation below

Step-by-step explanation:

a)

The combined resistance R is a function of the the two  

parallel  resistances

$\bf \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{R_1+R_2}{R_1R_2}$

So

$\bf R(R_1,R_2)=\frac{R_1R_2}{R_1+R_2}$

and we have

$\bf \frac{\partial R}{\partial R_1}=\frac{R_2(R_1+R_2)-R_1R_2}{(R_1+R_2)^2}=\frac{R_2^2}{(R_1+R_2)^2}$

Similarly,

$\bf \frac{\partial R}{\partial R_2}=\frac{R_1(R_1+R_2)-R_1R_2}{(R_1+R_2)^2}=\frac{R_1^2}{(R_1+R_2)^2}$

b)

If we divide both the numerator and denominator by $\bf R_1$ in the expression for R, we get

$\bf R=\frac{R_2}{1+R_2/R_1}$

hence, if we held $\bf R_2$ constant and increase $\bf R_1$ the fraction $\bf \frac{R_2}{R_1}$ gets smaller and so does the denominator of R, as a consequence R gets larger.

When $\bf R_1$ is very large, the denominator of R is close to 1, so R is close to $\bf R_2$

c)

By a symmetric reasoning, we see that R gets larger when holding $\bf R_1$ constant and $\bf R_2$ increases.

In this case, R gets closer to $\bf R_1$ as $\bf R_2$ grows.