The values of two resistors are `R_(1)=(6+-0.3)kOmega` and `R_(2)=(10+-0.2)kOmega`. The percentage error in the equivalent resistance when they are connected in parallel is

Options

A. 0.05125
B. 0.02
C. 0.03125
D. 0.10125

Explanation

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

### Step 1: Identify the values and uncertainties of the resistors
Given:
- \( R_1 = 6 \, \text{k}\Omega \) with an uncertainty of \( \Delta R_1 = 0.3 \, \text{k}\Omega \)
- \( R_2 = 10 \, \text{k}\Omega \) with an uncertainty of \( \Delta R_2 = 0.2 \, \text{k}\Omega \)

### Step 2: Calculate the equivalent resistance for resistors in parallel
The formula for the equivalent resistance \( R_{eq} \) of two resistors in parallel is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
This can be rearranged to:
\[ R_{eq} = \frac{R_1 R_2}{R_1 + R_2} \]

### Step 3: Substitute the values into the formula
Substituting the values of \( R_1 \) and \( R_2 \):
\[ R_{eq} = \frac{(6 \, \text{k}\Omega)(10 \, \text{k}\Omega)}{6 \, \text{k}\Omega + 10 \, \text{k}\Omega} = \frac{60 \, \text{k}\Omega^2}{16 \, \text{k}\Omega} = 3.75 \, \text{k}\Omega \]

### Step 4: Calculate the percentage error in \( R_{eq} \)
The percentage error in the equivalent resistance can be calculated using the formula for the propagation of uncertainty. For two resistors in parallel, the formula for the percentage error in \( R_{eq} \) is:
\[ \frac{\Delta R_{eq}}{R_{eq}} = \frac{\Delta R_1}{R_1} + \frac{\Delta R_2}{R_2} + \frac{\Delta (R_1 + R_2)}{R_1 + R_2} \]

### Step 5: Calculate each term
1. Calculate \( \frac{\Delta R_1}{R_1} \):
\[ \frac{\Delta R_1}{R_1} = \frac{0.3 \, \text{k}\Omega}{6 \, \text{k}\Omega} = 0.05 \]

2. Calculate \( \frac{\Delta R_2}{R_2} \):
\[ \frac{\Delta R_2}{R_2} = \frac{0.2 \, \text{k}\Omega}{10 \, \text{k}\Omega} = 0.02 \]

3. Calculate \( \frac{\Delta (R_1 + R_2)}{R_1 + R_2} \):
\[ R_1 + R_2 = 6 \, \text{k}\Omega + 10 \, \text{k}\Omega = 16 \, \text{k}\Omega \]
\[ \Delta (R_1 + R_2) = \Delta R_1 + \Delta R_2 = 0.3 \, \text{k}\Omega + 0.2 \, \text{k}\Omega = 0.5 \, \text{k}\Omega \]
\[ \frac{\Delta (R_1 + R_2)}{R_1 + R_2} = \frac{0.5 \, \text{k}\Omega}{16 \, \text{k}\Omega} = 0.03125 \]

### Step 6: Combine the errors
Now, combine the errors:
\[ \frac{\Delta R_{eq}}{R_{eq}} = 0.05 + 0.02 + 0.03125 = 0.10125 \]

### Step 7: Calculate the percentage error
To find the percentage error:
\[ \text{Percentage Error} = \frac{\Delta R_{eq}}{R_{eq}} \times 100 = 0.10125 \times 100 = 10.125\% \]

### Final Result
The percentage error in the equivalent resistance when the resistors are connected in parallel is approximately **10.13%**.