A 80 `Omega` resistor a 2 H inductor and a `5.07xx10^(-6)` F capacitor are connected in series to an AC power supply of 220 V, 50 Hz. Calculate the potential difference across the resistor.

Answer

To solve the problem of finding the potential difference across the resistor in a series LCR circuit connected to an AC power supply, we will follow these steps:

### Step 1: Identify Given Values
- Resistance (R) = 80 Ω
- Inductance (L) = 2 H
- Capacitance (C) = 5.07 x 10^(-6) F
- Voltage (V_rms) = 220 V
- Frequency (f) = 50 Hz

### Step 2: Calculate the Inductive Reactance (X_L)
The formula for inductive reactance is:
\[ X_L = \omega L = 2 \pi f L \]
Substituting the values:
\[ X_L = 2 \pi (50) (2) = 628.32 \, \Omega \]

### Step 3: Calculate the Capacitive Reactance (X_C)
The formula for capacitive reactance is:
\[ X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C} \]
Substituting the values:
\[ X_C = \frac{1}{2 \pi (50) (5.07 \times 10^{-6})} \approx 628.32 \, \Omega \]

### Step 4: Determine the Impedance (Z) of the Circuit
In a series LCR circuit, the impedance (Z) can be calculated using the formula:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Since \(X_L\) and \(X_C\) are equal, we have:
\[ Z = \sqrt{80^2 + (628.32 - 628.32)^2} = \sqrt{80^2} = 80 \, \Omega \]

### Step 5: Calculate the Current (I) in the Circuit
Using Ohm's law for AC circuits:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
Substituting the values:
\[ I_{rms} = \frac{220}{80} = 2.75 \, A \]

### Step 6: Calculate the Potential Difference Across the Resistor (V_R)
Using Ohm's law:
\[ V_R = I_{rms} \times R \]
Substituting the values:
\[ V_R = 2.75 \times 80 = 220 \, V \]

### Final Answer
The potential difference across the resistor is **220 V**.

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