A coil has resistance `30 ohm` and inductive reactance `20 ohm`at `50 Hz` frequency. If an ac source of 200 volts. `100 Hz`, is connected across the coil, the current in the coil will be

Options

A. `4.0 A`
B. `8.0 A`
C. `(20)/(sqrt12) A`
D. `2.0 A`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Resistance (R) = 30 ohms
- Inductive reactance at 50 Hz (X_L) = 20 ohms
- Voltage (V) = 200 volts
- New frequency (f) = 100 Hz

### Step 2: Calculate the inductance (L) of the coil
Using the formula for inductive reactance:
\[ X_L = 2 \pi f L \]
At 50 Hz, we have:
\[ 20 = 2 \pi (50) L \]
Rearranging gives:
\[ L = \frac{20}{100 \pi} = \frac{1}{5 \pi} \text{ H} \]

### Step 3: Calculate the new inductive reactance (X_L) at 100 Hz
Using the same formula for inductive reactance:
\[ X_L = 2 \pi f L \]
At 100 Hz:
\[ X_L = 2 \pi (100) \left(\frac{1}{5 \pi}\right) \]
This simplifies to:
\[ X_L = 40 \text{ ohms} \]

### Step 4: Calculate the total impedance (Z)
The total impedance in an R-L circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the values:
\[ Z = \sqrt{30^2 + 40^2} \]
Calculating gives:
\[ Z = \sqrt{900 + 1600} = \sqrt{2500} = 50 \text{ ohms} \]

### Step 5: Calculate the current (I_rms)
Using Ohm's law for AC circuits:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
Substituting the values:
\[ I_{rms} = \frac{200}{50} = 4 \text{ A} \]

### Final Answer
The current in the coil is **4 A**.

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