A bulb is connected to a 100 V DC supply . It is found that the current of 5 A flows in the circuit. The same bulb is connected to an AC supply at 120 V . The frequency of source is 50 Hz. Calculate the inductance of coil required so that the lamp glows in an AC circuit too.

Answer

To solve the problem step by step, we will follow the logical sequence of calculations based on the information provided in the question.

### Step 1: Calculate the Resistance of the Bulb
Given that the bulb operates at 100 V DC with a current of 5 A, we can calculate the resistance (R) of the bulb using Ohm's Law.

\[ R = \frac{V}{I} = \frac{100 \, \text{V}}{5 \, \text{A}} = 20 \, \Omega \]

### Step 2: Understand the AC Circuit Configuration
In the AC circuit, the bulb will be connected in series with an inductor (L). We need to ensure that the voltage drop across the bulb is 100 V when connected to a 120 V AC supply.

### Step 3: Calculate the Impedance of the Circuit
The total impedance (Z) in the circuit, which consists of the resistance (R) and the inductive reactance (X_L), is given by:

\[ Z = \sqrt{R^2 + X_L^2} \]

### Step 4: Calculate the RMS Current in the Circuit
The RMS current (I_RMS) flowing through the circuit can be expressed in terms of the RMS voltage (V_RMS) and the impedance (Z):

\[ I_{RMS} = \frac{V_{RMS}}{Z} = \frac{120 \, \text{V}}{Z} \]

### Step 5: Calculate the Voltage Drop Across the Bulb
The voltage drop across the bulb (V_bulb) can be expressed as:

\[ V_{bulb} = I_{RMS} \cdot R \]

We want this voltage drop to be equal to 100 V:

\[ 100 \, \text{V} = I_{RMS} \cdot 20 \, \Omega \]

### Step 6: Substitute I_RMS into the Voltage Equation
Substituting the expression for I_RMS into the voltage equation gives:

\[ 100 = \left(\frac{120}{Z}\right) \cdot 20 \]

### Step 7: Solve for Z
Rearranging the equation to solve for Z:

\[ Z = \frac{120 \cdot 20}{100} = 24 \, \Omega \]

### Step 8: Relate Impedance to Resistance and Inductive Reactance
Now we can relate the impedance back to the resistance and inductive reactance:

\[ Z^2 = R^2 + X_L^2 \]

Substituting the known values:

\[ 24^2 = 20^2 + X_L^2 \]

Calculating:

\[ 576 = 400 + X_L^2 \]
\[ X_L^2 = 576 - 400 = 176 \]
\[ X_L = \sqrt{176} \approx 13.3 \, \Omega \]

### Step 9: Relate Inductive Reactance to Inductance
Inductive reactance (X_L) is related to inductance (L) and frequency (f) by the formula:

\[ X_L = \omega L = 2 \pi f L \]

Substituting for X_L:

\[ 13.3 = 2 \pi (50) L \]

### Step 10: Solve for Inductance L
Rearranging to solve for L:

\[ L = \frac{13.3}{2 \pi (50)} \approx \frac{13.3}{314.16} \approx 0.0424 \, \text{H} \, \text{or} \, 42.4 \, \text{mH} \]

### Final Answer
The inductance required for the coil is approximately:

\[ L \approx 0.0424 \, \text{H} \, \text{or} \, 42.4 \, \text{mH} \]

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