1> A coil has resistance 10 ohm and an inductance 0.4 H. It is connected to an AC source of 6.5V, 30/pi Hz. Find average power consumed in the circuit.

Ans: 0.625 W ( I m getting 1.625 W)

2> A bulb rated 60W 220V is connected across a household supply of AC of 220V. Calculate maximum instantaneous current through the filament.

Ans: .039 A

To tackle these questions, we need to apply some fundamental concepts from electrical engineering, specifically related to alternating current (AC) circuits and power calculations. Let’s break down each problem step by step.

1. Average Power Consumed in an AC Circuit

We have a coil with a resistance (R) of 10 ohms and an inductance (L) of 0.4 H connected to an AC source of 6.5 V at a frequency of 30/π Hz. To find the average power consumed in the circuit, we first need to determine the impedance (Z) of the coil.

Calculating Impedance

The impedance in an R-L circuit can be calculated using the formula:

• Z = √(R² + (X_L)²)

Where X_L is the inductive reactance, calculated as:

• X_L = 2πfL

Substituting the values:

• f = 30/π Hz
• L = 0.4 H

Calculating X_L:

• X_L = 2π(30/π)(0.4) = 24 Ω

Now, we can find the total impedance:

• Z = √(10² + 24²) = √(100 + 576) = √676 = 26 Ω

Finding Current

Next, we calculate the current (I) flowing through the circuit using Ohm's Law:

• I = V/Z

Substituting the values:

• I = 6.5 V / 26 Ω = 0.25 A

Calculating Average Power

The average power (P) consumed in an AC circuit is given by:

• P = I²R

Substituting the current and resistance:

• P = (0.25 A)² × 10 Ω = 0.625 W

This matches the answer you provided, confirming that the average power consumed in the circuit is indeed 0.625 W.

2. Maximum Instantaneous Current Through a Bulb

Now, let’s move on to the second question regarding a bulb rated at 60 W and 220 V connected to a household supply of 220 V. We want to find the maximum instantaneous current through the filament.

Understanding Power and Voltage

The power rating of the bulb (P) is given as 60 W, and the voltage (V) is 220 V. The relationship between power, voltage, and current (I) is given by:

• P = VI

Calculating Current

Rearranging the formula to find the current:

• I = P/V

Substituting the values:

• I = 60 W / 220 V = 0.2727 A

Finding Maximum Instantaneous Current

In an AC circuit, the maximum instantaneous current (I_max) can be calculated using the RMS (Root Mean Square) value of the current. The RMS value is what we calculated above. The maximum instantaneous current can be found using:

• I_max = I_RMS × √2

Substituting the RMS current:

• I_max = 0.2727 A × √2 ≈ 0.385 A

However, if we are looking for the current through the filament at the rated voltage, we can directly use the RMS current value, which is approximately 0.2727 A. If you are looking for the maximum instantaneous current, it would be around 0.385 A, but the average current through the filament is what you might be referring to, which is about 0.2727 A.

In summary, the maximum instantaneous current through the filament is approximately 0.385 A, while the average current is about 0.2727 A. If you have any further questions or need clarification, feel free to ask!