If potential `V = 100 +- 0.5` Volt and current `I = 10 +- 0.2` amp are given to us. Then what will be the value of resistance

Options

A. `10 +- 0.7` ohm
B. `5 +- 2` ohm
C. `0.1 +- 0.2` ohm
D. None of these

Answer

To find the resistance \( R \) given the potential \( V \) and current \( I \) along with their uncertainties, we can follow these steps:

### Step 1: Identify the given values
We are given:
- Potential \( V = 100 \pm 0.5 \) Volts
- Current \( I = 10 \pm 0.2 \) Amperes

### Step 2: Calculate the resistance without considering the error
Using Ohm's Law, the resistance \( R \) can be calculated as:
\[ R = \frac{V}{I} \]
Substituting the values:
\[ R = \frac{100 \, \text{V}}{10 \, \text{A}} = 10 \, \Omega \]

### Step 3: Calculate the error in resistance
To find the error in resistance \( \Delta R \), we use the formula for propagation of uncertainty:
\[ \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \]
Where:
- \( \Delta V = 0.5 \, \text{V} \)
- \( V = 100 \, \text{V} \)
- \( \Delta I = 0.2 \, \text{A} \)
- \( I = 10 \, \text{A} \)

### Step 4: Substitute the values into the error formula
Now substituting the values:
\[ \frac{\Delta R}{10} = \frac{0.5}{100} + \frac{0.2}{10} \]

Calculating each term:
\[ \frac{0.5}{100} = 0.005 \]
\[ \frac{0.2}{10} = 0.02 \]

Adding these together:
\[ \frac{\Delta R}{10} = 0.005 + 0.02 = 0.025 \]

### Step 5: Solve for \( \Delta R \)
Now, multiply both sides by 10 to find \( \Delta R \):
\[ \Delta R = 10 \times 0.025 = 0.25 \, \Omega \]

### Step 6: Write the final result for resistance including error
Thus, the final value of resistance with its uncertainty is:
\[ R = 10 \pm 0.25 \, \Omega \]

### Summary
The resistance is \( 10 \, \Omega \) with an uncertainty of \( 0.25 \, \Omega \).

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