A coil a capacitor and an `AC` source of rms voltage `24 V` are connected in series. By varying the frequency of the source, a maximum rms current of 6 A is observed. If coil is connected is at DC batteryof emf 12 volt and internal resistance `4Omega`, then current through it in steady state is

Options

A. 2.4 A
B. 1.8 A
C. 1.5 A
D. 1.2 A

Explanation

To solve the problem step by step, we will analyze the given information and apply the relevant formulas.

### Step 1: Understand the AC Circuit
We have a coil (inductor) and a capacitor connected in series with an AC source. The RMS voltage of the AC source is given as \( V_{\text{rms}} = 24 \, V \) and the maximum RMS current observed is \( I_{\text{rms}} = 6 \, A \).

### Step 2: Calculate the Resistance in the AC Circuit
At resonance in an RLC circuit, the impedance \( Z \) is equal to the resistance \( R \). We can find \( R \) using the formula:
\[ R = \frac{V_{\text{rms}}}{I_{\text{rms}}} \]
Substituting the given values:
\[ R = \frac{24 \, V}{6 \, A} = 4 \, \Omega \]

### Step 3: Analyze the DC Circuit
Now, we need to find the current through the coil when it is connected to a DC battery with an EMF of \( 12 \, V \) and an internal resistance of \( 4 \, \Omega \).

### Step 4: Calculate the Total Resistance in the DC Circuit
The total resistance \( R_{\text{total}} \) in the DC circuit is the sum of the resistance of the coil (which we found to be \( 4 \, \Omega \)) and the internal resistance of the battery:
\[ R_{\text{total}} = R + R_{\text{internal}} = 4 \, \Omega + 4 \, \Omega = 8 \, \Omega \]

### Step 5: Calculate the Current in the Steady State
Using Ohm's law, the current \( I \) through the coil in the steady state can be calculated as:
\[ I = \frac{V}{R_{\text{total}}} \]
Substituting the values:
\[ I = \frac{12 \, V}{8 \, \Omega} = 1.5 \, A \]

### Final Answer
The current through the coil in steady state when connected to the DC battery is \( 1.5 \, A \).

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