Two nonideal batteries are connected in parallel. Consider the following statements:
(A)The equivalent emf is smaller than either of the two emfs.
(B) The equivalent internal resistance is smaller than either of the two internal resistances.

Options

A. The equivalent emf is smaller than either of the two emfs
B. The equivalent5 internal resistance is smaler than either of the two internal resistances
C. B is correct b ut A is wrong
D. both A are B are wrong

Answer

To solve the problem regarding two non-ideal batteries connected in parallel, we will analyze the equivalent emf and equivalent internal resistance of the system.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have two non-ideal batteries connected in parallel.
- Let the emf of the first battery be \( E_1 \) and its internal resistance be \( R_1 \).
- Let the emf of the second battery be \( E_2 \) and its internal resistance be \( R_2 \).

2. **Finding the Equivalent Emf**:
- The formula for the equivalent emf \( E_{eq} \) when two batteries are connected in parallel is given by:
\[ E_{eq} = \frac{E_1/R_1 + E_2/R_2}{1/R_1 + 1/R_2} \]
- This expression indicates that the equivalent emf depends on both the emfs and the internal resistances of the batteries.

3. **Special Case**:
- If both batteries have the same emf (i.e., \( E_1 = E_2 = E \)), the equivalent emf simplifies to:
\[ E_{eq} = E \cdot \frac{R_1 + R_2}{R_1 + R_2} = E \]
- Thus, in this special case, the equivalent emf is equal to the emf of a single cell.

4. **General Case**:
- In the general case where \( E_1 \) and \( E_2 \) are not equal, the equivalent emf \( E_{eq} \) will still be less than the larger of the two emfs due to the division by the sum of the resistances in the denominator.
- Therefore, we can conclude that the equivalent emf is smaller than either of the two emfs.

5. **Finding the Equivalent Internal Resistance**:
- The equivalent internal resistance \( R_{eq} \) when batteries are connected in parallel is given by:
\[ R_{eq} = \frac{R_1 R_2}{R_1 + R_2} \]
- This formula shows that the equivalent internal resistance is always less than either \( R_1 \) or \( R_2 \) (as the product of resistances divided by their sum is always less than the smallest resistance).

6. **Conclusion**:
- From the analysis:
- Statement (A): The equivalent emf is smaller than either of the two emfs. **True**
- Statement (B): The equivalent internal resistance is smaller than either of the two internal resistances. **True**

### Final Answer:
Both statements (A) and (B) are correct.