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. a) Both A and B are correct
B. b) A is correct but B is wrong
C. c) B is correct but A is wrong
D. d) Both A and B are incorrect

Answer

To solve the problem regarding two non-ideal batteries connected in parallel, we need to analyze the statements given about their equivalent emf and internal resistance.

### Step 1: Understanding the Equivalent EMF
When two batteries with emfs \(E_1\) and \(E_2\) and internal resistances \(R_1\) and \(R_2\) are connected in parallel, the equivalent emf (\(E_{eq}\)) can be calculated using the formula:

\[ E_{eq} = \frac{E_1 R_2 + E_2 R_1}{R_1 + R_2} \]

If we assume that both batteries are identical, meaning \(E_1 = E_2 = E\) and \(R_1 = R_2 = R\), then:

\[ E_{eq} = \frac{E R + E R}{R + R} = \frac{2E R}{2R} = E \]

Thus, the equivalent emf is equal to the emf of either battery, not smaller than either of the two emfs. Therefore, statement (A) is **incorrect**.

### Step 2: Understanding the Equivalent Internal Resistance
Next, we calculate the equivalent internal resistance (\(R_{eq}\)) for the two batteries in parallel. The formula for equivalent resistance in parallel is:

\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]

For identical batteries:

\[ \frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \implies R_{eq} = \frac{R}{2} \]

Since \(R_{eq} = \frac{R}{2}\), the equivalent internal resistance is smaller than the internal resistance of either battery (\(R\)). Therefore, statement (B) is **correct**.

### Conclusion
- Statement (A) is incorrect: The equivalent emf is not smaller than either of the two emfs.
- Statement (B) is correct: The equivalent internal resistance is smaller than either of the two internal resistances.

Thus, the correct answer is option **C**: B is correct, but A is wrong.

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