If alternating current of 60 Hz frequency is flowing through inductance of L = 1 mH and drop in `DeltaV_(L)` is 0.6 V then alternating current-

Options

A. `(1)/(pi) A `
B. `(5)/(pi)` A
C. `(50)/(pi) ` A
D. `(20)/(pi)` A

Answer

To solve the problem of finding the alternating current (I) flowing through an inductor with a given inductance (L) and voltage drop (ΔV_L), we can follow these steps:

### Step 1: Identify Given Values
- Frequency (f) = 60 Hz
- Inductance (L) = 1 mH = 1 × 10^(-3) H
- Voltage drop across the inductor (ΔV_L) = 0.6 V

### Step 2: Calculate the Inductive Reactance (X_L)
The inductive reactance (X_L) can be calculated using the formula:
\[ X_L = \omega L \]
where \( \omega = 2\pi f \).

First, calculate \( \omega \):
\[ \omega = 2\pi \times 60 \, \text{Hz} = 120\pi \, \text{rad/s} \]

Now, substitute the values into the formula for X_L:
\[ X_L = 120\pi \times (1 \times 10^{-3}) = 0.12\pi \, \text{ohms} \]

### Step 3: Calculate the Alternating Current (I)
Using Ohm's law for AC circuits, the current (I) can be calculated as:
\[ I = \frac{\Delta V_L}{X_L} \]

Substituting the values we have:
\[ I = \frac{0.6 \, \text{V}}{0.12\pi \, \text{ohms}} \]

### Step 4: Simplify the Expression
Calculating the current:
\[ I = \frac{0.6}{0.12\pi} = \frac{6}{1.2\pi} = \frac{5}{\pi} \, \text{A} \]

### Final Answer
The alternating current is:
\[ I = \frac{5}{\pi} \, \text{A} \]

---