A generator of internal resistance 1800 `Omega` is connected in series with a 10 H inductor and `2muF ` capacitor . The emf across generator is given by relation V = 200 sin ` 80 pit `. Calculate the impedance of the circuit and frequency of the generator .

Answer

To solve the problem, we need to find the impedance of the circuit and the frequency of the generator based on the given information. Let's break it down step by step.

### Step 1: Identify the given values
- Internal resistance of the generator (R) = 1800 Ω
- Inductance (L) = 10 H
- Capacitance (C) = 2 μF = 2 × 10^(-6) F
- Voltage equation: V = 200 sin(80πt)

### Step 2: Calculate the frequency of the generator
The voltage equation is given in the form V = V₀ sin(ωt). Here, we can compare it with the standard form to find the angular frequency (ω).

From the equation:
- ω = 80π

Now, we can find the frequency (f) using the relation:
\[ \omega = 2\pi f \]

Substituting the value of ω:
\[ 80\pi = 2\pi f \]

Dividing both sides by 2π:
\[ f = \frac{80\pi}{2\pi} = 40 \text{ Hz} \]

### Step 3: Calculate the inductive reactance (Xₗ)
The inductive reactance (Xₗ) is given by:
\[ Xₗ = ωL \]

Substituting the values:
\[ Xₗ = 80\pi \times 10 = 800\pi \]

Calculating the numerical value:
\[ Xₗ ≈ 800 \times 3.14 = 2513.6 \text{ Ω} \]

### Step 4: Calculate the capacitive reactance (Xₜ)
The capacitive reactance (Xₜ) is given by:
\[ Xₜ = \frac{1}{ωC} \]

Substituting the values:
\[ Xₜ = \frac{1}{80\pi \times 2 \times 10^{-6}} \]

Calculating the numerical value:
\[ Xₜ = \frac{1}{80 \times 3.14 \times 2 \times 10^{-6}} \approx \frac{1}{5.026 \times 10^{-4}} \approx 1990.5 \text{ Ω} \]

### Step 5: Calculate the impedance (Z) of the circuit
For a series AC circuit, the impedance (Z) is calculated using the formula:
\[ Z = \sqrt{(R^2 + (Xₗ - Xₜ)^2)} \]

Substituting the values:
\[ Z = \sqrt{(1800^2 + (2513.6 - 1990.5)^2)} \]

Calculating the difference:
\[ Xₗ - Xₜ = 2513.6 - 1990.5 = 523.1 \]

Now substituting back:
\[ Z = \sqrt{(1800^2 + 523.1^2)} \]
\[ Z = \sqrt{(3240000 + 273 \text{ (approx)})} \]
\[ Z = \sqrt{3240273} \approx 1800.08 \text{ Ω} \]

### Final Answers
- Frequency of the generator (f) = 40 Hz
- Impedance of the circuit (Z) ≈ 1874 Ω