Respuesta :
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 [tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]
To find :
a) Approximate the change in R ?
b) Compute the actual change.
Solution :
a) Approximate the change in R
[tex]R_1=12\ ohm[/tex] and [tex]R_2=10\ ohm[/tex]
[tex]R_1[/tex] is decreased from 12 ohms to 11 ohms.
i.e. [tex]\triangle R_1=21-11=1\ ohm[/tex]
[tex]R_2[/tex] is increased from 10 ohms to 11 ohms.
i.e. [tex]\triangle R_2=11-10=1\ ohm[/tex]
The change in R is given by,
[tex]\frac{1}{\triangle R}=\frac{1}{\triangle R_1}+\frac{1}{\triangle R_2}[/tex]
[tex]\frac{1}{\triangle R}=\frac{\triangle R_2+\triangle R_1}{(\triangle R_1)(\triangle R_2)}[/tex]
[tex]\triangle R=\frac{(\triangle R_1)(\triangle R_2)}{\triangle R_2+\triangle R_1}[/tex]
[tex]\triangle R=\frac{(1)(1)}{1+1}[/tex]
[tex]\triangle R=\frac{1}{2}[/tex]
[tex]\triangle R=0.5\ ohm[/tex]
b) The actual change in Resistance
[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]
[tex]\frac{1}{R}=\frac{1}{12}+\frac{1}{10}[/tex]
[tex]R=\frac{10\times 12}{10+12}[/tex]
[tex]R=\frac{120}{22}[/tex]
[tex]R=5.46\ ohm[/tex]
When resistances are charged, [tex]R_1=R_2=11[/tex]
[tex]\frac{1}{R'}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]
[tex]\frac{1}{R'}=\frac{1}{11}+\frac{1}{11}[/tex]
[tex]R'=\frac{11}{2}[/tex]
[tex]R'=5.5\ ohm[/tex]
Change in resistance is given by,
[tex]C=R'-R[/tex]
[tex]C=5.5-5.46[/tex]
[tex]C=0.04\ ohm[/tex]
