An inductor coil and a capacitor and a resistance of `5Omega` are connected in series to an a.c. source of rms voltage 30 V. When the frequency of the source is varied, a maximum r.m.s. current of 5 A is observed. If this inductor is connected in parallel with a resistance `5Omega` to a battery of emf 25 V and internal resistance `2.0Omega`, the current drawn from the battery is

Options

A. `(150)/(17)A`
B. `2.5 A`
C. `(125)/(33)A`
D. `(125)/(99)A`

Answer

To solve the problem, we need to find the current drawn from the battery when the inductor is connected in parallel with a resistor. Here’s a step-by-step solution:

### Step 1: Determine the internal resistance of the inductor
From the given information, we know that the maximum RMS current (I_max) is 5 A when the RMS voltage (V_rms) is 30 V. The formula for RMS current in an AC circuit is:

\[ I_{rms} = \frac{V_{rms}}{Z} \]

Where Z is the impedance of the circuit. At maximum current, the impedance Z is minimized, which occurs at resonance when:

\[ Z = R + r \]

Given that \(I_{rms} = 5 \, \text{A}\) and \(V_{rms} = 30 \, \text{V}\), we can write:

\[ 5 = \frac{30}{R + r} \]

Rearranging gives:

\[ R + r = \frac{30}{5} = 6 \, \Omega \]

### Step 2: Identify the resistance values
We are given that the resistance \(R\) of the resistor is \(5 \, \Omega\). Therefore, we can substitute this into our equation:

\[ 5 + r = 6 \]

Solving for \(r\):

\[ r = 6 - 5 = 1 \, \Omega \]

So, the internal resistance of the inductor is \(1 \, \Omega\).

### Step 3: Analyze the new circuit configuration
Now, we need to find the current drawn from a battery of emf \(25 \, V\) with an internal resistance of \(2 \, \Omega\) when the inductor (with internal resistance \(1 \, \Omega\)) is connected in parallel with a resistor of \(5 \, \Omega\).

### Step 4: Calculate the equivalent resistance of the inductor and resistor in parallel
The equivalent resistance \(R_{eq}\) of the inductor and the \(5 \, \Omega\) resistor in parallel is given by:

\[ \frac{1}{R_{eq}} = \frac{1}{r} + \frac{1}{R} \]

Substituting the values we found:

\[ \frac{1}{R_{eq}} = \frac{1}{1} + \frac{1}{5} \]

Calculating this gives:

\[ \frac{1}{R_{eq}} = 1 + 0.2 = 1.2 \]

Thus:

\[ R_{eq} = \frac{1}{1.2} = \frac{5}{6} \, \Omega \]

### Step 5: Find the total resistance in the circuit
The total resistance \(R_{total}\) in the circuit, which includes the internal resistance of the battery, is:

\[ R_{total} = R_{eq} + R_{internal} = \frac{5}{6} + 2 \]

Converting \(2\) into sixths gives:

\[ R_{total} = \frac{5}{6} + \frac{12}{6} = \frac{17}{6} \, \Omega \]

### Step 6: Calculate the current drawn from the battery
Using Ohm's law, the current \(I\) drawn from the battery can be calculated as:

\[ I = \frac{V}{R_{total}} = \frac{25}{\frac{17}{6}} = 25 \times \frac{6}{17} = \frac{150}{17} \, \text{A} \]

### Final Answer
The current drawn from the battery is:

\[ I = \frac{150}{17} \, \text{A} \]