Question

The total resistance R of two resistors connected in parallel circuit is given by 1/R = 1/R_1 + 1/R_2. Approximate the change in R as R_1 is decreased from 12 ohms to 11 ohms and R_2 is increased from 10 ohms to 11 ohms. Compute the actual change.

Answer 1

a) Approximate the change in R is 0.5 ohm.

b) The actual change in R is 0.04 ohm.

Explanation:

Given : The total resistance R of two resistors connected in parallel circuit is given by
`(1)/(R)=(1)/(R_1)+(1)/(R_2)`

To find :

a) Approximate the change in R ?

b) Compute the actual change.

Solution :

a) Approximate the change in R

`R_1=12\ ohm` and
`R_2=10\ ohm`

`R_1` is decreased from 12 ohms to 11 ohms.

i.e.
`\triangle R_1=21-11=1\ ohm`

`R_2` is increased from 10 ohms to 11 ohms.

i.e.
`\triangle R_2=11-10=1\ ohm`

The change in R is given by,

`(1)/(\triangle R)=(1)/(\triangle R_1)+(1)/(\triangle R_2)`

`(1)/(\triangle R)=(\triangle R_2+\triangle R_1)/((\triangle R_1)(\triangle R_2))`

`\triangle R=((\triangle R_1)(\triangle R_2))/(\triangle R_2+\triangle R_1)`

`\triangle R=((1)(1))/(1+1)`

`\triangle R=(1)/(2)`

`\triangle R=0.5\ ohm`

b) The actual change in Resistance

`(1)/(R)=(1)/(R_1)+(1)/(R_2)`

`(1)/(R)=(1)/(12)+(1)/(10)`

`R=(10* 12)/(10+12)`

`R=(120)/(22)`

`R=5.46\ ohm`

When resistances are charged,
`R_1=R_2=11`

`(1)/(R')=(1)/(R_1)+(1)/(R_2)`

`(1)/(R')=(1)/(11)+(1)/(11)`

`R'=(11)/(2)`

`R'=5.5\ ohm`

Change in resistance is given by,

`C=R'-R`

`C=5.5-5.46`

`C=0.04\ ohm`