Ques Regarding A.C.Circuits

An electric lamp which runs 100V dc & comsumes 10A current is connected to ac mains at 150V, 50Hz cycles with a choke coil in series. Calculate the inductance and drop of voltage across the choke. Neglect resistance of choke.

(Please give step-by-step solution)

To solve this problem, we need to analyze the circuit involving an electric lamp and a choke coil when connected to an AC mains supply. We will calculate the inductance of the choke coil and the voltage drop across it. Let's break this down step by step.

Understanding the Circuit

The lamp is rated for 100V DC and consumes 10A of current. When connected to an AC supply of 150V at 50Hz, we need to ensure that the lamp operates safely. The choke coil is used to limit the current flowing through the lamp by introducing inductance into the circuit.

Step 1: Calculate the Power Rating of the Lamp

The power consumed by the lamp can be calculated using the formula:

• P = V × I

Substituting the values:

• P = 100V × 10A = 1000W

Step 2: Determine the Equivalent AC Voltage for the Lamp

When the lamp is connected to the AC supply, we need to consider the effective voltage. Since the lamp is rated for 100V DC, we will use this as the reference for the AC voltage. The lamp will effectively operate at an RMS voltage of 100V.

Step 3: Calculate the Total Impedance in the Circuit

The total voltage supplied is 150V, and the voltage across the lamp is 100V. Therefore, the voltage drop across the choke coil can be calculated as:

• V_choke = V_supply - V_lamp

Substituting the values:

• V_choke = 150V - 100V = 50V

Step 4: Calculate the Current in the Circuit

Since the lamp consumes 10A, the current flowing through the choke coil is also 10A (assuming series connection).

Step 5: Calculate the Inductive Reactance

The inductive reactance (X_L) can be calculated using the formula:

• X_L = V_choke / I

Substituting the values:

• X_L = 50V / 10A = 5Ω

Step 6: Calculate the Inductance of the Choke Coil

The inductive reactance is also related to the inductance (L) and the frequency (f) by the formula:

• X_L = 2πfL

Rearranging this gives us:

• L = X_L / (2πf)

Substituting the values (with f = 50Hz):

• L = 5Ω / (2π × 50Hz)
• L ≈ 5 / 314.16 ≈ 0.0159 H or 15.9 mH

Summary of Results

In summary, the calculations yield:

• Inductance of the choke coil: 15.9 mH
• Voltage drop across the choke: 50V

This analysis shows how the choke coil limits the current and allows the lamp to operate safely within its rated voltage when connected to an AC supply. If you have any further questions or need clarification on any of the steps, feel free to ask!