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

a)

`\bf (\partial R)/(\partial R_1)=(R_2^2)/((R_1+R_2)^2)`

`\bf (\partial R)/(\partial R_2)=(R_1^2)/((R_1+R_2)^2)`

b) and c)

See explanation below

Explanation:

a)

The combined resistance R is a function of the the two

parallel resistances

`\bf (1)/(R)=(1)/(R_1)+(1)/(R_2)=(R_1+R_2)/(R_1R_2)`

So

`\bf R(R_1,R_2)=(R_1R_2)/(R_1+R_2)`

and we have

`\bf (\partial R)/(\partial R_1)=(R_2(R_1+R_2)-R_1R_2)/((R_1+R_2)^2)=(R_2^2)/((R_1+R_2)^2)`

Similarly,

`\bf (\partial R)/(\partial R_2)=(R_1(R_1+R_2)-R_1R_2)/((R_1+R_2)^2)=(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=(R_2)/(1+R_2/R_1)`

hence, if we held
`\bf R_2` constant and increase
`\bf R_1` the fraction
`\bf (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.