A `50 Omega` resistor an inductor and a capacitor are connected to an AC source 220 V. The frequency of source is 50 Hz. Calculate the current in the circuit if voltage across resistor capacitor and inductor is 220 V, 260 V and 260 V, respectively.

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the Given Values
- Resistance (R) = 50 Ω
- Voltage across the resistor (VR) = 220 V
- Voltage across the capacitor (VC) = 260 V
- Voltage across the inductor (VL) = 260 V
- RMS Voltage (V_RMS) = 220 V
- Frequency (f) = 50 Hz

### Step 2: Understand the Circuit
In an AC circuit with a resistor (R), inductor (L), and capacitor (C) connected in series, the total voltage is the vector sum of the voltages across each component. The voltage across the resistor is in phase with the current, while the voltages across the inductor and capacitor are out of phase with the current.

### Step 3: Use the Voltage Relationship
From the given information, we can see that:
- The voltage across the inductor (VL) and the voltage across the capacitor (VC) are equal (260 V).

Since these two voltages are equal and opposite in phase, we can say:
\[ V_L = V_C \]

### Step 4: Determine the Reactances
Using Ohm's law for AC circuits:
- The inductive reactance \( X_L \) can be expressed as:
\[ V_L = I \cdot X_L \]
- The capacitive reactance \( X_C \) can be expressed as:
\[ V_C = I \cdot X_C \]

Since \( V_L = V_C \), we have:
\[ I \cdot X_L = I \cdot X_C \]
This implies:
\[ X_L = X_C \]

### Step 5: Calculate the Impedance
In a series RLC circuit, the total impedance \( Z \) can be calculated using:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]

Since \( X_L = X_C \), we can simplify this to:
\[ Z = R \]

### Step 6: Calculate the Current
Now, we can find the current \( I \) using the formula:
\[ I = \frac{V_{RMS}}{Z} \]
Substituting the values:
\[ I = \frac{220 V}{50 Ω} \]
\[ I = 4.4 A \]

### Final Answer
The current in the circuit is **4.4 A**.

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