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7nadin3 [17]
3 years ago
5

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.
Mathematics
1 answer:
Naily [24]3 years ago
3 0

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

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