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

Answer

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

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

### Step 2: Calculate the net reactance (X)
The net reactance (X) of the circuit can be calculated using the formula:
\[ X = XL - XC \]
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:
\[ Z = \sqrt{121 + 49} = \sqrt{170} \]
Approximating:
\[ Z \approx 13.04 Ω \]

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

### Step 5: Round the current to appropriate significant figures
Rounding the result:
\[ I \approx 20 A \]

### Final Answer
The current through the circuit is approximately **20 amperes**.

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