A circuit consists of a resistance of 20Ω, an inductance of 0.05H, connected in

To solve this problem, we need to use the formulas and concepts from electrical circuit theory. Let's go step by step to find the required information:

i) Impedance (Z):
The impedance of a series RL circuit is given by the formula:
Z = √(R^2 + (ωL)^2)
where R is the resistance, ω is the angular frequency, and L is the inductance.

In this case, R = 20Ω and L = 0.05H. The angular frequency is calculated as ω = 2πf, where f is the frequency in Hz. Given that the frequency is 50Hz, ω = 2π(50) = 100π.

By substituting these values into the formula, we get:
Z = √(20^2 + (100π*0.05)^2)
= √(400 + 500π^2)
≈ 53.14Ω (rounded to 2 decimal places)

ii) Current (I):
Using Ohm's Law, we can calculate the current (I) as:
I = V / Z
where V is the applied voltage.

In this case, V = 230V. By substituting V and Z into the formula, we have:
I = 230 / 53.14
≈ 4.33A (rounded to 2 decimal places)

iii) Power Factor (PF):
The power factor (PF) is given by the formula:
PF = cos(Φ)
where Φ is the phase angle between the voltage and current in the circuit.

In a purely resistive circuit, the phase angle is 0°, and the power factor is 1. However, in a circuit with inductance, there will be a phase difference between the voltage and current.

In this case, since the circuit contains an inductance, the power factor will be less than 1.

iv) Power Consumed (P):
The power consumed (P) by the circuit can be calculated using the formula:
P = V * I * PF
where V is the applied voltage, I is the current, and PF is the power factor.

By substituting the given values into the formula, we obtain:
P = 230 * 4.33 * PF

v) Voltage Drop across R and L:
The voltage drop across a resistor (VR) in a series circuit can be calculated using Ohm's Law as VR = I * R.

The voltage drop across an inductor (VL) can be calculated using the formula VL = I * ωL.

vi) Vector Diagram:
To draw the vector diagram, we need to represent the applied voltage (V) and the current (I) on a complex plane. The angle between them represents the phase difference, and the magnitude represents the amplitude.

I hope this helps you in finding the answers to the given questions about the RL circuit.