Calculate equivalent resistance of two resistors `R_(1) and R_(2)` in parallel where, `R_(1) = (6+-0.2)`ohm and ` R_2 = (3+-0.1)ohm`

Explanation

To calculate the equivalent resistance of two resistors \( R_1 \) and \( R_2 \) in parallel, we will follow these steps:

### Step 1: Identify the given values
We have:
- \( R_1 = 6 \pm 0.2 \, \text{ohm} \)
- \( R_2 = 3 \pm 0.1 \, \text{ohm} \)

### Step 2: Determine the range of values for \( R_1 \) and \( R_2 \)
The range for \( R_1 \):
- Minimum: \( 6 - 0.2 = 5.8 \, \text{ohm} \)
- Maximum: \( 6 + 0.2 = 6.2 \, \text{ohm} \)

The range for \( R_2 \):
- Minimum: \( 3 - 0.1 = 2.9 \, \text{ohm} \)
- Maximum: \( 3 + 0.1 = 3.1 \, \text{ohm} \)

### Step 3: Use the formula for equivalent resistance in parallel
The formula for the equivalent resistance \( R_{\text{eq}} \) of two resistors in parallel is given by:
\[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \]
This can also be expressed as:
\[ R_{\text{eq}} = \frac{R_1 \cdot R_2}{R_1 + R_2} \]

### Step 4: Calculate the mean value of \( R_{\text{eq}} \)
Substituting the mean values of \( R_1 \) and \( R_2 \):
\[ R_{\text{eq}} = \frac{6 \cdot 3}{6 + 3} = \frac{18}{9} = 2 \, \text{ohm} \]

### Step 5: Calculate the uncertainty in \( R_{\text{eq}} \)
To find the uncertainty \( \Delta R \), we will use the formula for the propagation of errors in parallel resistors:
\[ \frac{\Delta R}{R_{\text{eq}}^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2} \]
Where:
- \( \Delta R_1 = 0.2 \, \text{ohm} \)
- \( \Delta R_2 = 0.1 \, \text{ohm} \)

Substituting the values:
\[ \frac{\Delta R}{2^2} = \frac{0.2}{6^2} + \frac{0.1}{3^2} \]
Calculating each term:
\[ \frac{\Delta R}{4} = \frac{0.2}{36} + \frac{0.1}{9} \]
\[ \frac{\Delta R}{4} = \frac{0.2}{36} + \frac{0.1 \cdot 4}{36} = \frac{0.2 + 0.4}{36} = \frac{0.6}{36} = \frac{1}{60} \]
Now, multiplying both sides by 4:
\[ \Delta R = 4 \cdot \frac{1}{60} = \frac{4}{60} = \frac{1}{15} \approx 0.0667 \, \text{ohm} \]

### Step 6: Final result
The final equivalent resistance with its uncertainty is:
\[ R_{\text{eq}} = 2 \pm 0.07 \, \text{ohm} \]