It is known to all of you that the impedance of a circuit is dependent on the frequency of source. In order to study the effect of frequency on the impedance, a student in a lab took 2 impedance boxes `P` and `Q` and connected them in series with an `AC` source of variable frequency. The emf of the source is constant at `10 V` Box `P` contains a capacitance of `1muF` in series with a resistance of `32 Omega`. And the box `Q` has a coil of self-inductance `4.9 mH` and a resistance of `68 Omega`in series. He adjusted the frequency so that the maximum current flows in `P` and `Q`. Based on his experimental set up and the reading by him at various moment, answer the following questions.
The angular frequency foer which he detects maximum current in the circuit is

Options

A. `10^5/7rad//s`
B. `10^4rad//s`
C. `10^5rad//s`
D. `10^4/7rad//s`

Explanation

To find the angular frequency at which the maximum current flows in the circuit, we can follow these steps:

### Step 1: Understand the concept of resonance
In an RLC circuit (which consists of resistance, inductance, and capacitance), maximum current occurs at resonance. The condition for resonance in such a circuit is given by the formula:
\[ \omega = \frac{1}{\sqrt{LC}} \]
where \( \omega \) is the angular frequency, \( L \) is the inductance, and \( C \) is the capacitance.

### Step 2: Identify the values of L and C
From the problem statement:
- The inductance \( L \) in box \( Q \) is given as \( 4.9 \, \text{mH} = 4.9 \times 10^{-3} \, \text{H} \).
- The capacitance \( C \) in box \( P \) is given as \( 1 \, \mu\text{F} = 1 \times 10^{-6} \, \text{F} \).

### Step 3: Substitute the values into the resonance formula
Now, we can substitute the values of \( L \) and \( C \) into the resonance formula:
\[ \omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(4.9 \times 10^{-3}) \times (1 \times 10^{-6})}} \]

### Step 4: Calculate the product of L and C
First, calculate the product \( LC \):
\[ LC = 4.9 \times 10^{-3} \times 1 \times 10^{-6} = 4.9 \times 10^{-9} \]

### Step 5: Calculate the square root
Now, find the square root:
\[ \sqrt{LC} = \sqrt{4.9 \times 10^{-9}} = \sqrt{4.9} \times 10^{-4.5} \approx 2.2136 \times 10^{-4.5} \]
Calculating \( \sqrt{4.9} \) gives approximately \( 2.2136 \).

### Step 6: Calculate the angular frequency
Now substituting back into the equation for \( \omega \):
\[ \omega = \frac{1}{\sqrt{4.9 \times 10^{-9}}} = \frac{1}{2.2136 \times 10^{-4.5}} \approx \frac{1}{4.9 \times 10^{-5}} = \frac{10^{5}}{4.9} \]

### Step 7: Final calculation
Calculating the final value gives:
\[ \omega \approx \frac{10^{5}}{4.9} \approx 20408.16 \, \text{rad/s} \]

### Conclusion
Thus, the angular frequency for which the student detects maximum current in the circuit is approximately:
\[ \omega \approx 20408.16 \, \text{rad/s} \]