Two resistance are measured in ohm and is given as:-
`R_(1)=3Omega+-1% &R_(2)=6Omega+-2%`
When they are connected in parallel, the percentage error in equivalent resistance is

Options

A. 0.03
B. `4.5%`
C. `0.67%`
D. `1.33%`

Explanation

To find the percentage error in the equivalent resistance when two resistors are connected in parallel, we can follow these steps:

### Step 1: Understand the given values and errors
We have:
- \( R_1 = 3 \, \Omega \pm 1\% \)
- \( R_2 = 6 \, \Omega \pm 2\% \)

### Step 2: Calculate the equivalent resistance without error
For resistors in parallel, the equivalent resistance \( R_{eq} \) is given by the formula:
\[ R_{eq} = \frac{R_1 R_2}{R_1 + R_2} \]
Substituting the values:
\[ R_{eq} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 \, \Omega \]

### Step 3: Determine the relative errors
The relative error for \( R_1 \) and \( R_2 \) can be calculated as:
- For \( R_1 \):
\[ \text{Relative error in } R_1 = \frac{1\%}{100} = 0.01 \]
- For \( R_2 \):
\[ \text{Relative error in } R_2 = \frac{2\%}{100} = 0.02 \]

### Step 4: Use the formula for error in equivalent resistance
The formula for the relative error in the equivalent resistance \( R_{eq} \) when resistors are in parallel is:
\[ \frac{D R_{eq}}{R_{eq}} = \frac{D R_1}{R_1} + \frac{D R_2}{R_2} \]
Substituting the values:
\[ \frac{D R_{eq}}{R_{eq}} = \frac{0.01}{1} + \frac{0.02}{1} = 0.01 + 0.02 = 0.03 \]

### Step 5: Calculate the percentage error
To find the percentage error in \( R_{eq} \):
\[ \text{Percentage error} = \frac{D R_{eq}}{R_{eq}} \times 100 = 0.03 \times 100 = 3\% \]

### Final Result
The percentage error in the equivalent resistance when \( R_1 \) and \( R_2 \) are connected in parallel is **3%**.

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