An inductive circuit contains a resistance of 10 ohms and an inductance of 2 henry. If an alternating voltage of 120 V and frequency 60 Hz is applied to this circuit, the current in the circuit would be nearly

Options

A. 0.32 A
B. 0.80 A
C. 0.48 A
D. 0.16 A

Answer

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

### Step 1: Identify the given values
- Resistance (R) = 10 ohms
- Inductance (L) = 2 henry
- Frequency (f) = 60 Hz
- Voltage (V) = 120 V

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the value of frequency:
\[ \omega = 2\pi \times 60 = 120\pi \text{ rad/s} \]

### Step 3: Calculate the inductive reactance (X_L)
The inductive reactance (X_L) is given by the formula:
\[ X_L = \omega L \]
Substituting the values of ω and L:
\[ X_L = 120\pi \times 2 = 240\pi \text{ ohms} \]

### Step 4: Calculate the impedance (Z)
The impedance (Z) in an inductive circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the values of R and X_L:
\[ Z = \sqrt{10^2 + (240\pi)^2} \]
Calculating \(10^2\):
\[ 10^2 = 100 \]
Calculating \((240\pi)^2\):
\[ (240\pi)^2 = 57600\pi^2 \]
Now, substituting these values into the impedance formula:
\[ Z = \sqrt{100 + 57600\pi^2} \]

### Step 5: Calculate the numerical value of Z
Using the approximate value of \(\pi \approx 3.14\):
\[ 57600\pi^2 \approx 57600 \times (3.14)^2 \approx 57600 \times 9.8596 \approx 567576 \]
Now, adding this to 100:
\[ Z \approx \sqrt{567576 + 100} = \sqrt{567676} \approx 754.5 \text{ ohms} \]

### Step 6: Calculate the current (I)
Using Ohm's law, the current (I) can be calculated as:
\[ I = \frac{V}{Z} \]
Substituting the values of V and Z:
\[ I = \frac{120}{754.5} \approx 0.159 \text{ A} \]

### Final Answer
The current in the circuit would be nearly **0.159 A**.