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.
Impedance of box P at the above frequency is

Options

A. `70Omega`
B. `77 Omega`
C. `90 Omega`
D. `100 Omega`

Explanation

To find the impedance of box P at the frequency where maximum current flows, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Components of Box P:**
- Box P contains a capacitance \( C = 1 \mu F = 1 \times 10^{-6} F \) in series with a resistance \( R_P = 32 \Omega \).

2. **Determine the Condition for Maximum Current:**
- Maximum current in an AC circuit occurs when the circuit is at resonance. At resonance, the capacitive reactance \( X_C \) and inductive reactance \( X_L \) are equal.

3. **Calculate the Capacitive Reactance \( X_C \):**
- The capacitive reactance is given by the formula:
\[ X_C = \frac{1}{\omega C} \]
- To find \( \omega \) (angular frequency), we need to use the relationship at resonance, which is:
\[ \omega = \frac{1}{\sqrt{LC}} \]
- However, we do not have the inductance \( L \) for box P, but we can still calculate \( X_C \) directly using the frequency that gives maximum current.

4. **Determine the Frequency \( f \):**
- From the video, it is stated that the frequency is related to the inductance of box Q. The inductance \( L = 4.9 mH = 4.9 \times 10^{-3} H \).
- The resonant frequency \( f \) can be calculated as:
\[ f = \frac{1}{2\pi \sqrt{LC}} \]
- Substituting \( L \) and \( C \):
\[ f = \frac{1}{2\pi \sqrt{4.9 \times 10^{-3} \times 1 \times 10^{-6}}} \]

5. **Calculate \( \omega \):**
- Convert frequency to angular frequency:
\[ \omega = 2\pi f \]

6. **Calculate \( X_C \):**
- Substitute \( \omega \) back into the capacitive reactance formula:
\[ X_C = \frac{1}{\omega C} \]

7. **Calculate the Impedance \( Z_P \):**
- The impedance of box P is given by:
\[ Z_P = \sqrt{R_P^2 + X_C^2} \]
- Substitute \( R_P \) and \( X_C \) to find \( Z_P \).

8. **Final Calculation:**
- After performing the calculations, you will find \( Z_P \).

### Final Answer:
The impedance of box P is approximately \( Z_P \approx 77 \Omega \).

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