The potential difference across the terminals of a battery is `8.4 V` when there is a current of `1.50 A` in the battery from the negative to the positive terminal. When the current is `3.50 Air`, the reverse direction, the potential difference becomes `9.4 V`.
(a) What is the internal resistance of the battery?
(b) What is the emf of the battery?

Explanation

To solve the problem, we will use the concepts of internal resistance and electromotive force (emf) of the battery.

### Given:
1. When current \( I_1 = 1.50 \, \text{A} \) (discharging), potential difference \( V_1 = 8.4 \, \text{V} \).
2. When current \( I_2 = 3.50 \, \text{A} \) (charging), potential difference \( V_2 = 9.4 \, \text{V} \).

### (a) Finding the internal resistance of the battery:

1. **Using the formula for discharging:**
\[ V_1 = E - I_1 R \]
Rearranging gives:
\[ E = V_1 + I_1 R \]
Substituting the values:
\[ E = 8.4 + 1.5R \quad \text{(1)} \]

2. **Using the formula for charging:**
\[ V_2 = E + I_2 R \]
Rearranging gives:
\[ E = V_2 - I_2 R \]
Substituting the values:
\[ E = 9.4 - 3.5R \quad \text{(2)} \]

3. **Equating the two expressions for \( E \):**
\[ 8.4 + 1.5R = 9.4 - 3.5R \]

4. **Solving for \( R \):**
\[ 8.4 + 1.5R + 3.5R = 9.4 \]
\[ 5R = 9.4 - 8.4 \]
\[ 5R = 1 \]
\[ R = \frac{1}{5} = 0.2 \, \Omega \]

### (b) Finding the emf of the battery:

1. **Using the value of \( R \) in either equation (1) or (2):**
Using equation (1):
\[ E = 8.4 + 1.5(0.2) \]
\[ E = 8.4 + 0.3 \]
\[ E = 8.7 \, \text{V} \]

### Final Answers:
- (a) The internal resistance of the battery is \( R = 0.2 \, \Omega \).
- (b) The emf of the battery is \( E = 8.7 \, \text{V} \).