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

Options

A. `(20)/sqrt(13)` A
B. 2.0 A
C. 4.0 A
D. 8.0 A

Answer

To solve the problem step-by-step, we will follow the outlined process in the video transcript.

### Step 1: Identify the given values
- Resistance (R) = 30 Ω
- Inductive reactance at 50 Hz (X_L) = 20 Ω
- Frequency of AC source (f) = 100 Hz
- Voltage (V) = 200 V

### Step 2: Calculate the inductance (L)
We know that the inductive reactance (X_L) is given by the formula:
\[ X_L = \omega L \]
where \( \omega = 2 \pi f \).

At 50 Hz:
\[ \omega = 2 \pi \times 50 = 100 \pi \]

Now, substituting the values into the equation for X_L:
\[ 20 = 100 \pi L \]

To find 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
Now, we will calculate the new inductive reactance when the frequency is changed to 100 Hz.

At 100 Hz:
\[ \omega' = 2 \pi \times 100 = 200 \pi \]

Using the inductance we found:
\[ X_L' = \omega' L = 200 \pi \times \frac{1}{5 \pi} \]

Simplifying this:
\[ X_L' = 40 \, \Omega \]

### Step 4: Calculate the impedance (Z)
The impedance (Z) in an R-L circuit is given by:
\[ Z = \sqrt{R^2 + X_L'^2} \]

Substituting the values:
\[ Z = \sqrt{30^2 + 40^2} \]
\[ Z = \sqrt{900 + 1600} \]
\[ Z = \sqrt{2500} \]
\[ Z = 50 \, \Omega \]

### Step 5: Calculate the current (I)
Using Ohm's law, the current can be calculated as:
\[ I = \frac{V}{Z} \]

Substituting the values:
\[ I = \frac{200}{50} \]
\[ I = 4 \, \text{A} \]

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

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