Two resistor of resistance `R_(1)=(6+-0.09)Omega` and `R_(2)=(3+-0.09)Omega` are connected in parallel the equivalent resistance R with error (in `Omega)`

Options

A. `R=(2+-0.04)`
B. `R=(2+-0.05)`
C. `R=(9+-0.18)`
D. `R=(2+-0.18)`

Explanation

To find the equivalent resistance \( R \) of two resistors \( R_1 \) and \( R_2 \) connected in parallel, we can use the formula:

\[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \]

### Step 1: Substitute the values of \( R_1 \) and \( R_2 \)

Given:
- \( R_1 = 6 \, \Omega \)
- \( R_2 = 3 \, \Omega \)

Substituting these values into the formula:

\[ \frac{1}{R} = \frac{1}{6} + \frac{1}{3} \]

### Step 2: Calculate the right-hand side

To add the fractions, we need a common denominator. The least common multiple of 6 and 3 is 6.

\[ \frac{1}{3} = \frac{2}{6} \]

Now substituting back:

\[ \frac{1}{R} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2} \]

### Step 3: Find \( R \)

Taking the reciprocal of both sides gives:

\[ R = 2 \, \Omega \]

### Step 4: Calculate the error in \( R \)

To find the error in the equivalent resistance, we use the formula for the error in parallel resistors. The formula for the error in equivalent resistance \( R \) when resistors are in parallel is given by:

\[ \Delta R = R \cdot \sqrt{\left(\frac{\Delta R_1}{R_1}\right)^2 + \left(\frac{\Delta R_2}{R_2}\right)^2} \]

Where:
- \( \Delta R_1 = 0.09 \, \Omega \)
- \( \Delta R_2 = 0.09 \, \Omega \)

Substituting the known values:

\[ \Delta R = 2 \cdot \sqrt{\left(\frac{0.09}{6}\right)^2 + \left(\frac{0.09}{3}\right)^2} \]

### Step 5: Calculate the individual terms

Calculating the fractions:

\[ \frac{0.09}{6} = 0.015 \quad \text{and} \quad \frac{0.09}{3} = 0.03 \]

Now squaring these values:

\[ (0.015)^2 = 0.000225 \quad \text{and} \quad (0.03)^2 = 0.0009 \]

### Step 6: Sum the squares

Adding these squared values:

\[ 0.000225 + 0.0009 = 0.001125 \]

### Step 7: Take the square root

Now taking the square root:

\[ \sqrt{0.001125} \approx 0.0335 \]

### Step 8: Calculate the error

Now substituting back to find \( \Delta R \):

\[ \Delta R = 2 \cdot 0.0335 \approx 0.067 \]

### Final Result

Thus, the equivalent resistance \( R \) with error is:

\[ R = 2 \pm 0.067 \, \Omega \]