An inductor-coil , a capacitor and an AC source of rms voltage `24 V` are connected in series. When the frequency of the source is varied, a maximum rms current of `6.0 A` is observed. If this inductor coil is connected to a battery of `emf 12 V` and internal resistance `4.0 Omega`, what will be the current?

Explanation

To solve the problem step by step, we will follow the information provided in the question and the video transcript.

### Step 1: Understand the given values
- The RMS voltage of the AC source, \( E_{\text{rms}} = 24 \, \text{V} \)
- The maximum RMS current observed, \( I_{\text{rms}} = 6.0 \, \text{A} \)
- The EMF of the battery, \( E = 12 \, \text{V} \)
- The internal resistance of the battery, \( r = 4.0 \, \Omega \)

### Step 2: Calculate the total resistance in the AC circuit
Using Ohm's law, we know that the resistance \( R \) can be calculated from the RMS voltage and current:
\[ R = \frac{E_{\text{rms}}}{I_{\text{rms}}} \]
Substituting the given values:
\[ R = \frac{24 \, \text{V}}{6.0 \, \text{A}} = 4 \, \Omega \]

### Step 3: Calculate the total resistance when the inductor is connected to the battery
The total resistance in the circuit when the inductor coil is connected to the battery will be the sum of the internal resistance of the battery and the resistance of the inductor:
\[ R_{\text{total}} = R + r = 4 \, \Omega + 4 \, \Omega = 8 \, \Omega \]

### Step 4: Calculate the current when the inductor is connected to the battery
Using Ohm's law again, we can find the current \( I \) flowing through the circuit when connected to the battery:
\[ I = \frac{E}{R_{\text{total}}} \]
Substituting the values:
\[ I = \frac{12 \, \text{V}}{8 \, \Omega} = 1.5 \, \text{A} \]

### Final Answer
The current when the inductor coil is connected to the battery is \( 1.5 \, \text{A} \).

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