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. `3%`
B. `4.5%`
C. `0.67%`
D. `1.33%`

Answer

To find the percentage error in the equivalent resistance when two resistances \( R_1 \) and \( R_2 \) are connected in parallel, we can follow these steps:

### Step 1: Identify the values and errors
Given:
- \( R_1 = 3 \, \Omega \) with an error of \( \pm 1\% \)
- \( R_2 = 6 \, \Omega \) with an error of \( \pm 2\% \)

### Step 2: Calculate the absolute errors
- The absolute error in \( R_1 \):
\[ \Delta R_1 = 1\% \text{ of } 3 \, \Omega = 0.01 \times 3 = 0.03 \, \Omega \]

- The absolute error in \( R_2 \):
\[ \Delta R_2 = 2\% \text{ of } 6 \, \Omega = 0.02 \times 6 = 0.12 \, \Omega \]

### Step 3: Calculate the equivalent resistance
The formula for equivalent resistance \( R_{eq} \) when resistances are in parallel is:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
Thus,
\[ R_{eq} = \frac{R_1 \cdot R_2}{R_1 + R_2} \]
Substituting the values:
\[ R_{eq} = \frac{3 \cdot 6}{3 + 6} = \frac{18}{9} = 2 \, \Omega \]

### Step 4: Calculate the relative error in equivalent resistance
The relative error in \( R_{eq} \) when resistances are multiplied or divided is given by the sum of the relative errors:
\[ \text{Relative error} = \frac{\Delta R_1}{R_1} + \frac{\Delta R_2}{R_2} \]
Substituting the values:
\[ \text{Relative error} = \frac{0.03}{3} + \frac{0.12}{6} \]
Calculating each term:
\[ \frac{0.03}{3} = 0.01 \quad \text{and} \quad \frac{0.12}{6} = 0.02 \]
Thus,
\[ \text{Relative error} = 0.01 + 0.02 = 0.03 \]

### Step 5: Convert relative error to percentage error
To find the percentage error, we multiply the relative error by 100:
\[ \text{Percentage error} = 0.03 \times 100 = 3\% \]

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

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