A potentail difference of `V = (20 +-1)` volt is applied across a resistance of `(8.0 +-2)` ohm. Calculate the current with error limits.

Explanation

To solve the problem of calculating the current with error limits when a potential difference of \( V = (20 \pm 1) \) volts is applied across a resistance of \( R = (8.0 \pm 2) \) ohms, we will use Ohm's Law, which states:

\[ I = \frac{V}{R} \]

### Step 1: Calculate the nominal current
First, we calculate the nominal current using the central values of \( V \) and \( R \).

\[ I = \frac{20 \, \text{V}}{8.0 \, \Omega} = 2.5 \, \text{A} \]

### Step 2: Calculate the relative errors
Next, we need to calculate the relative errors in \( V \) and \( R \).

1. **Error in Voltage (\( \Delta V \))**:
\[ \Delta V = 1 \, \text{V} \]
The relative error in voltage is:
\[ \frac{\Delta V}{V} = \frac{1}{20} \]

2. **Error in Resistance (\( \Delta R \))**:
\[ \Delta R = 2.0 \, \Omega \]
The relative error in resistance is:
\[ \frac{\Delta R}{R} = \frac{2}{8} = \frac{1}{4} \]

### Step 3: Calculate the total relative error in current
Using the formula for the propagation of uncertainty, we can find the relative error in current \( I \):

\[ \frac{\Delta I}{I} = \frac{\Delta V}{V} + \frac{\Delta R}{R} \]

Substituting the values we calculated:

\[ \frac{\Delta I}{I} = \frac{1}{20} + \frac{1}{4} \]

To add these fractions, we need a common denominator. The least common multiple of 20 and 4 is 20.

\[ \frac{1}{20} + \frac{5}{20} = \frac{6}{20} = \frac{3}{10} \]

### Step 4: Calculate the absolute error in current
Now we can find the absolute error in current \( \Delta I \):

\[ \Delta I = I \times \frac{\Delta I}{I} = 2.5 \times \frac{3}{10} = 0.75 \, \text{A} \]

### Step 5: Write the final result with error limits
Finally, we can express the current with its error limits:

\[ I = 2.5 \pm 0.75 \, \text{A} \]

### Summary
The current calculated with error limits is:

\[ I = 2.5 \pm 0.75 \, \text{A} \]