The voltage measured across resistance is `V = (50 pm 0.1)V ` and current is` I = ( 10 pm 0.2)`. Find error in resistance

Options

A. `pm 1.12 ohm`
B. `pm 0.11 ohm`
C. `pm 0.21 ohm`
D. `pm 2.12 ohm`

Answer

To find the error in resistance given the voltage and current measurements, we can follow these steps:

### Step 1: Write down the given values
- Voltage, \( V = 50 \pm 0.1 \, V \)
- Current, \( I = 10 \pm 0.2 \, A \)

### Step 2: Calculate the nominal value of resistance
Resistance \( R \) is given by the formula:
\[ R = \frac{V}{I} \]
Substituting the nominal values:
\[ R_0 = \frac{50}{10} = 5 \, \Omega \]

### Step 3: Differentiate the resistance formula
To find the error in resistance \( \Delta R \), we differentiate the resistance formula:
\[ \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \]
where:
- \( \Delta V = 0.1 \, V \) (error in voltage)
- \( \Delta I = 0.2 \, A \) (error in current)

### Step 4: Substitute the values into the differentiation equation
Now we substitute the values into the differentiated equation:
\[ \Delta R = R_0 \left( \frac{\Delta V}{V} + \frac{\Delta I}{I} \right) \]
Substituting the known values:
\[ \Delta R = 5 \left( \frac{0.1}{50} + \frac{0.2}{10} \right) \]

### Step 5: Calculate each term
Calculating the first term:
\[ \frac{0.1}{50} = 0.002 \]
Calculating the second term:
\[ \frac{0.2}{10} = 0.02 \]

### Step 6: Add the terms
Now add these two results:
\[ \Delta R = 5 \left( 0.002 + 0.02 \right) = 5 \times 0.022 = 0.11 \, \Omega \]

### Step 7: Write the final result
The error in the measurement of resistance is:
\[ \Delta R = \pm 0.11 \, \Omega \]

Thus, the final answer is:
\[ R = 5 \pm 0.11 \, \Omega \]