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

Answer:

a) Approximate the change in R is 0.5 ohm.

b) The actual change in R is 0.04 ohm.

Step-by-step explanation:

Given : The total resistance R of two resistors connected in parallel circuit is given by $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

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,

$\frac{1}{\triangle R}=\frac{1}{\triangle R_1}+\frac{1}{\triangle R_2}$

$\frac{1}{\triangle R}=\frac{\triangle R_2+\triangle R_1}{(\triangle R_1)(\triangle R_2)}$

$\triangle R=\frac{(\triangle R_1)(\triangle R_2)}{\triangle R_2+\triangle R_1}$

$\triangle R=\frac{(1)(1)}{1+1}$

$\triangle R=\frac{1}{2}$

$\triangle R=0.5\ ohm$

b) The actual change in Resistance

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

$\frac{1}{R}=\frac{1}{12}+\frac{1}{10}$

$R=\frac{10\times 12}{10+12}$

$R=\frac{120}{22}$

$R=5.46\ ohm$

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

$\frac{1}{R'}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R'}=\frac{1}{11}+\frac{1}{11}$

$R'=\frac{11}{2}$

$R'=5.5\ ohm$

Change in resistance is given by,

$C=R'-R$

$C=5.5-5.46$

$C=0.04\ ohm$