An ammeter is connected to measure the current intensity in a circuit with a resistance R. What relative error will be made if connection of the ammeter does not change the current intensity in the circuit? The voltage across the ends of the circuit is kept constant.

Options

A. `(R )/(R_(0))`
B. `(R_(0))/(R )`
C. `(1)/(1+(R )/(R_(0)))`
D. `(1)/(1+(R_(0) )/(R ))`

Answer

To solve the problem, we need to analyze the situation where an ammeter is connected to a circuit with a resistance \( R \) and a constant voltage across the circuit. We will derive the relative error in the current measurement due to the ammeter's resistance.

### Step-by-Step Solution:

1. **Understand the relationship between current, voltage, and resistance**:
The current \( I \) in a circuit is given by Ohm's Law:
\[ I = \frac{V}{R} \]
where \( V \) is the voltage across the circuit and \( R \) is the resistance.

2. **Introduce the ammeter's resistance**:
When an ammeter with resistance \( R_0 \) is connected in series with the circuit, the total resistance becomes:
\[ R_{total} = R + R_0 \]

3. **Determine the new current with the ammeter connected**:
The current \( I' \) with the ammeter connected is given by:
\[ I' = \frac{V}{R + R_0} \]

4. **Calculate the relative error in current**:
The relative error in the current measurement can be expressed as:
\[ \text{Relative Error} = \frac{\Delta I}{I} \]
where \( \Delta I = I - I' \).

5. **Substituting the expressions for \( I \) and \( I' \)**:
We can express \( \Delta I \) as:
\[ \Delta I = I - I' = \frac{V}{R} - \frac{V}{R + R_0} \]
Factoring out \( V \):
\[ \Delta I = V \left( \frac{1}{R} - \frac{1}{R + R_0} \right) \]
Simplifying this gives:
\[ \Delta I = V \left( \frac{(R + R_0) - R}{R(R + R_0)} \right) = \frac{V R_0}{R(R + R_0)} \]

6. **Now, substitute \( I \) into the relative error formula**:
\[ \text{Relative Error} = \frac{\Delta I}{I} = \frac{\frac{V R_0}{R(R + R_0)}}{\frac{V}{R}} = \frac{R_0}{R + R_0} \]

7. **Rearranging the expression**:
This can be rewritten as:
\[ \text{Relative Error} = \frac{R_0}{R} \cdot \frac{1}{1 + \frac{R_0}{R}} = \frac{1}{1 + \frac{R}{R_0}} \]

8. **Final result**:
Thus, the relative error in the current measurement when the ammeter is connected is:
\[ \text{Relative Error} = \frac{1}{1 + \frac{R}{R_0}} \]

### Conclusion:
The correct answer is:
\[ \text{Relative Error} = \frac{1}{1 + \frac{R}{R_0}} \]