A Capacitor and a coil in series are connected to a 6volt ac source. By varying the frequency of the source, maximum current of 600mA is observed. If the same coil is now connected toa cell of emf 6volt dc and internal resistance of 2ohm, the current h through it will be

Options

A. 0.5A
B. 0.6A
C. 1.0A
D. 2.0A

Answer

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

### Step 1: Understanding the AC Circuit
We have a capacitor and a coil (inductor) connected in series to a 6V AC source. The maximum current observed is 600 mA when the frequency is varied.

### Step 2: Finding the Impedance at Resonance
At resonance, the inductive reactance (XL) equals the capacitive reactance (XC), and the impedance (Z) is minimized. The formula for impedance in a series RLC circuit is:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
At resonance, this simplifies to:
\[ Z = R \]

### Step 3: Using Maximum Current to Find Resistance
The maximum current (I_max) is given as 600 mA (or 0.6 A). The relationship between the maximum current, the RMS voltage (E_rms), and the impedance is:
\[ I_{max} = \frac{E_{rms}}{Z} \]
Substituting the known values:
\[ 0.6 = \frac{6}{R} \]
Rearranging gives:
\[ R = \frac{6}{0.6} = 10 \, \Omega \]

### Step 4: Analyzing the DC Circuit
Now, we connect the same coil to a 6V DC source with an internal resistance of 2Ω. The total resistance in the circuit will be the sum of the coil's resistance and the internal resistance:
\[ R_{total} = R + R_{internal} = 10 \, \Omega + 2 \, \Omega = 12 \, \Omega \]

### Step 5: Calculating the Current through the DC Circuit
Using Ohm's law, the current (I) through the circuit can be calculated as:
\[ I = \frac{E}{R_{total}} \]
Substituting the values:
\[ I = \frac{6}{12} = 0.5 \, A \]

### Final Answer
The current through the coil when connected to the 6V DC source is **0.5 A**.

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