Two resistance are given as `R_1 = (10 pm 0.5) Omaga` and `R_2 = (15 pm 0.5) Omaga` . The percentage error in the measurement of equivalent resistance when they are connected in parallel is

Options

A. 6.33
B. 2.33
C. 5.33
D. 4.33

Answer

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

### Step 1: Write down the values of resistances and their uncertainties
Given:
- \( R_1 = 10 \pm 0.5 \, \Omega \)
- \( R_2 = 15 \pm 0.5 \, \Omega \)

### Step 2: Use the formula for equivalent resistance in parallel
The formula for the equivalent resistance \( R \) when two resistances \( R_1 \) and \( R_2 \) are connected in parallel is:
\[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \]

### Step 3: Calculate the equivalent resistance
Substituting the values:
\[ \frac{1}{R} = \frac{1}{10} + \frac{1}{15} \]
Finding a common denominator (which is 30):
\[ \frac{1}{R} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} \]
Thus,
\[ R = \frac{30}{5} = 6 \, \Omega \]

### Step 4: Calculate the uncertainty in equivalent resistance
The formula for the uncertainty in \( R \) when resistances are in parallel is given by:
\[ \frac{\Delta R}{R} = \sqrt{\left(\frac{\Delta R_1}{R_1}\right)^2 + \left(\frac{\Delta R_2}{R_2}\right)^2} \]
Where \( \Delta R_1 = 0.5 \, \Omega \) and \( \Delta R_2 = 0.5 \, \Omega \).

### Step 5: Substitute the values into the uncertainty formula
Calculating the individual terms:
\[ \frac{\Delta R_1}{R_1} = \frac{0.5}{10} = 0.05 \]
\[ \frac{\Delta R_2}{R_2} = \frac{0.5}{15} \approx 0.0333 \]

Now substituting these into the uncertainty formula:
\[ \frac{\Delta R}{R} = \sqrt{(0.05)^2 + (0.0333)^2} \]
Calculating:
\[ \frac{\Delta R}{R} = \sqrt{0.0025 + 0.0011} = \sqrt{0.0036} \approx 0.06 \]

### Step 6: Calculate the absolute uncertainty
Now, calculate \( \Delta R \):
\[ \Delta R = R \times \frac{\Delta R}{R} = 6 \times 0.06 = 0.36 \, \Omega \]

### Step 7: Calculate the percentage error
The percentage error is given by:
\[ \text{Percentage Error} = \left(\frac{\Delta R}{R}\right) \times 100 = 0.06 \times 100 = 6\% \]

### Final Answer
The percentage error in the measurement of equivalent resistance when the two resistances are connected in parallel is approximately **6%**.

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