A circuit has a resistance of `11 Omega`, an inductive reactance of `25 Omega`, and a capacitive resistance of `18 Omega`. It is connected to an `AC` source of `260 V` and `50 Hz`. The current through the circuit (in amperes) is

Options

A. `11`
B. `15`
C. `18`
D. `20`

Explanation

To solve the problem, we need to find the current flowing through the circuit given the resistance, inductive reactance, capacitive reactance, and the voltage of the AC source. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Resistance (R) = 11 Ω
- Inductive Reactance (X_L) = 25 Ω
- Capacitive Reactance (X_C) = 18 Ω
- Voltage (V) = 260 V

### Step 2: Calculate the net reactance (X)
The net reactance (X) in the circuit can be calculated using the formula:
\[ X = X_L - X_C \]
Substituting the values:
\[ X = 25 Ω - 18 Ω = 7 Ω \]

### Step 3: Calculate the impedance (Z)
The impedance (Z) of the circuit can be calculated using the formula:
\[ Z = \sqrt{R^2 + X^2} \]
Substituting the values:
\[ Z = \sqrt{(11 Ω)^2 + (7 Ω)^2} \]
Calculating:
\[ Z = \sqrt{121 + 49} = \sqrt{170} \]
\[ Z \approx 13.04 Ω \]

### Step 4: Calculate the current (I)
Using Ohm's law for AC circuits, the current can be calculated as:
\[ I = \frac{V}{Z} \]
Substituting the values:
\[ I = \frac{260 V}{13.04 Ω} \]
Calculating:
\[ I \approx 19.96 A \]

### Conclusion
The current through the circuit is approximately **20 A**.

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