An inductor coil, a capacitor and an AC source of rms voltage `24V` are connected in series. When the frequency of the source is varied, a maximam rms current of `6.0A` is observed. If this inductor coil is connected to a battery of emf `12V` and internal resistance `4.0 ohm`, then find the current (in ampere) flowing through the battery.

Answer

To solve the problem, we need to determine the current flowing through the battery when the inductor coil is connected to it. Given the parameters, we can use Ohm's law to find the current.

### Step-by-Step Solution:

1. **Identify the given values:**
- EMF of the battery (V) = 12V
- Internal resistance of the battery (r) = 4Ω

2. **Use Ohm's Law:**
Ohm's Law states that \( I = \frac{V}{R} \), where \( I \) is the current, \( V \) is the voltage, and \( R \) is the total resistance in the circuit.

3. **Calculate the current:**
Here, the total resistance in the circuit is just the internal resistance of the battery since there are no other resistances mentioned in the problem. Therefore, we can substitute the values into Ohm's Law:
\[ I = \frac{V}{r} = \frac{12V}{4Ω} = 3A \]

4. **Conclusion:**
The current flowing through the battery is \( 3A \).

### Final Answer:
The current flowing through the battery is **3 A**.