The possible error in computing the parallel resistance R of three resistances `1/R=(1)/R_1+ 1/R_2 + 1/R_3` from the formula , if `R_1, R_2, R_3` are each in error by plus 1.2% is :

Options

A. 0.012
B. 0.013
C. 0.014
D. 0.015

Answer

To find the possible error in computing the parallel resistance \( R \) of three resistances \( R_1, R_2, R_3 \) given that each resistance has an error of \( +1.2\% \), we will follow these steps:

### Step 1: Write the formula for parallel resistance
The formula for the total resistance \( R \) in parallel is given by:
\[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]

### Step 2: Differentiate the equation
To find the possible error, we differentiate the equation with respect to \( R \):
\[ d\left(\frac{1}{R}\right) = d\left(\frac{1}{R_1}\right) + d\left(\frac{1}{R_2}\right) + d\left(\frac{1}{R_3}\right) \]
Using the chain rule, we have:
\[ -\frac{dR}{R^2} = -\frac{dR_1}{R_1^2} - \frac{dR_2}{R_2^2} - \frac{dR_3}{R_3^2} \]

### Step 3: Rearranging the equation
Rearranging gives us:
\[ \frac{dR}{R} = \frac{dR_1}{R_1} + \frac{dR_2}{R_2} + \frac{dR_3}{R_3} \]

### Step 4: Expressing errors as percentages
Given that the errors in \( R_1, R_2, R_3 \) are \( +1.2\% \), we can express this as:
\[ \frac{dR_1}{R_1} \times 100 = 1.2, \quad \frac{dR_2}{R_2} \times 100 = 1.2, \quad \frac{dR_3}{R_3} \times 100 = 1.2 \]
Thus, we have:
\[ \frac{dR_1}{R_1} = 0.012, \quad \frac{dR_2}{R_2} = 0.012, \quad \frac{dR_3}{R_3} = 0.012 \]

### Step 5: Substitute the errors into the equation
Substituting these values into our rearranged equation:
\[ \frac{dR}{R} = 0.012 + 0.012 + 0.012 = 0.036 \]

### Step 6: Calculate the percentage error in \( R \)
To find the percentage error in \( R \):
\[ \frac{dR}{R} \times 100 = 0.036 \times 100 = 3.6\% \]

### Conclusion
The possible error in computing the parallel resistance \( R \) is \( 3.6\% \).

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