Which of the following circuits gives the correct value of resistance, when computed by using `R = (V//I)` where V and I are voltmeter and ammeter reading, respectively? The meters are not ideal.

Options

A.
B.
C.
D. none of these

Explanation

To solve the problem of determining which circuit gives the correct value of resistance using the formula \( R = \frac{V}{I} \), where \( V \) is the voltmeter reading and \( I \) is the ammeter reading, we need to analyze the behavior of the ammeter and voltmeter in different circuit configurations.

### Step-by-Step Solution:

1. **Understand the Concept**:
- The formula \( R = \frac{V}{I} \) gives the correct resistance value only if \( V \) is the potential difference across the resistor and \( I \) is the current flowing through that same resistor.

2. **Consider Non-Ideal Meters**:
- Since the meters are not ideal, they have finite resistances. Let \( R_a \) be the resistance of the ammeter and \( R_v \) be the resistance of the voltmeter. This means that both meters will affect the circuit operation.

3. **Analyze Circuit Cases**:
- **Case 1**: If the voltmeter is connected in parallel with the resistor:
- The voltmeter will draw some current \( I' \) due to its resistance \( R_v \).
- The current flowing through the resistor will then be \( I - I' \) (where \( I \) is the total current).
- The voltmeter will read the voltage across both the resistor and its own resistance, which will not give the correct voltage across the resistor alone.
- Thus, \( R = \frac{V}{I} \) will not yield the correct resistance.

- **Case 2**: If the ammeter is connected in series with the resistor:
- The ammeter will read the total current \( I \) flowing through the circuit.
- However, the voltmeter will measure the voltage drop across both the resistor and the ammeter's resistance \( R_a \).
- Therefore, the voltmeter reading \( V \) will not be the voltage across the resistor alone, leading to \( R = \frac{V}{I} \) not giving the correct resistance.

- **Case 3**: If the ammeter is connected in parallel with the resistor:
- The ammeter will not measure the current flowing through the resistor, as it is bypassed.
- This configuration is incorrect for measuring resistance since the ammeter will read a different current than what flows through the resistor.

4. **Conclusion**:
- In all analyzed cases, the configurations do not yield the correct resistance value when using \( R = \frac{V}{I} \) due to the non-ideal nature of the meters and their placements in the circuit.
- Therefore, the answer is that none of the circuits provide the correct value of resistance.