Give, potential difference V = (8 + 0.5) V and current I = (2 + 0.2) A. The value of resistance R is

Options

A. `(4 pm 16.25%)`
B. `(4pm 6.25%)`
C. `(4 pm 10%)`
D. `(4pm 8%)`

Answer

To solve the problem, we will use Ohm's law, which states that the resistance \( R \) is equal to the potential difference \( V \) divided by the current \( I \). We will also calculate the percentage error in the resistance based on the errors in the measurements of \( V \) and \( I \).

### Step-by-Step Solution:

1. **Identify the given values:**
- Potential difference \( V = 8 + 0.5 \) V (where \( 8 \) is the measured value and \( 0.5 \) is the uncertainty)
- Current \( I = 2 + 0.2 \) A (where \( 2 \) is the measured value and \( 0.2 \) is the uncertainty)

2. **Calculate the nominal value of resistance \( R \):**
- Using Ohm's law:
\[ R = \frac{V}{I} \]
- Substitute the nominal values:
\[ R = \frac{8}{2} = 4 \, \text{Ohms} \]

3. **Calculate the relative errors:**
- The relative error in \( V \) is given by:
\[ \frac{\Delta V}{V} = \frac{0.5}{8} \]
- The relative error in \( I \) is given by:
\[ \frac{\Delta I}{I} = \frac{0.2}{2} \]

4. **Calculate the total relative error in \( R \):**
- According to the formula for combining errors:
\[ \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \]
- Substitute the calculated relative errors:
\[ \frac{\Delta R}{R} = \frac{0.5}{8} + \frac{0.2}{2} \]

5. **Calculate the individual terms:**
- For \( \frac{0.5}{8} \):
\[ \frac{0.5}{8} = 0.0625 \]
- For \( \frac{0.2}{2} \):
\[ \frac{0.2}{2} = 0.1 \]

6. **Combine the relative errors:**
- Add the two relative errors:
\[ \frac{\Delta R}{R} = 0.0625 + 0.1 = 0.1625 \]

7. **Convert to percentage error:**
- To find the percentage error:
\[ \text{Percentage error in } R = 0.1625 \times 100\% = 16.25\% \]

8. **Final result for resistance with uncertainty:**
- Combine the nominal value of \( R \) with its uncertainty:
\[ R = 4 \pm 16.25\% \]

### Final Answer:
The value of resistance \( R \) is \( 4 \, \text{Ohms} \pm 16.25\% \).

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