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.