A series RLC circuit is connected to an AC supply 220 V, 50 Hz. The value of R is 10 `Omega` . The value of capacitive reactance of capacitor and incuctive reactance of inductor are 15 `Omega` and `20 Omega` respectively . Calculate the value of current in the circuit.

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the given values
- Voltage (V) = 220 V
- Frequency (f) = 50 Hz
- Resistance (R) = 10 Ω
- Capacitive Reactance (Xc) = 15 Ω
- Inductive Reactance (Xl) = 20 Ω

### Step 2: Calculate the net reactance (X)
The net reactance (X) in a series RLC circuit is given by the formula:
\[ X = Xl - Xc \]
Substituting the values:
\[ X = 20 \, \Omega - 15 \, \Omega = 5 \, \Omega \]

### Step 3: Calculate the impedance (Z)
The total impedance (Z) in a series RLC circuit can be calculated using the formula:
\[ Z = \sqrt{R^2 + X^2} \]
Substituting the values:
\[ Z = \sqrt{(10 \, \Omega)^2 + (5 \, \Omega)^2} \]
\[ Z = \sqrt{100 + 25} \]
\[ Z = \sqrt{125} \]
\[ Z = 5\sqrt{5} \, \Omega \]

### Step 4: Calculate the current (I)
The current (I) in the circuit can be calculated using Ohm's law for AC circuits:
\[ I = \frac{V}{Z} \]
Substituting the values:
\[ I = \frac{220 \, V}{5\sqrt{5} \, \Omega} \]

Calculating the current:
First, we need to calculate \( 5\sqrt{5} \):
\[ \sqrt{5} \approx 2.236 \]
Thus,
\[ 5\sqrt{5} \approx 5 \times 2.236 = 11.18 \, \Omega \]

Now substituting this back into the current formula:
\[ I \approx \frac{220}{11.18} \approx 19.67 \, A \]

### Final Answer
The value of the current in the circuit is approximately **19.67 A**.

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