Question

Consider a circuit in which a coil having inductance of 0.02H and resistance of 5Ω connected across 200V, 50Hz supply in series with a resistance of 5Ω. Considering all parameters with their behavior determine following below: Evaluate capacitive reactance of the circuit. Calculate inductive reactance of the circuit. Determine power factor of the circuit. Find total impedance of the circuit.​

Answer 1

Answer:

However, the analysis of a parallel RLC circuits can be a little more mathematically difficult than for series RLC circuits so in this tutorial about parallel RLC circuits only pure components are assumed in this tutorial to keep things simple.

This time instead of the current being common to the circuit components, the applied voltage is now common to all so we need to find the individual branch currents through each element. The total impedance, Z of a parallel RLC circuit is calculated using the current of the circuit similar to that for a DC parallel circuit, the difference this time is that admittance is used instead of impedance. Consider the parallel RLC circuit below.

Answer 2

Given:

A series LR circuit with

Inductor = 0.02H

Resistance= 5Ω

Supply voltage= 200V, 50Hz

To find:

1. Capacitive Reactance
2. Inductive Reactance
3. Impedance
4. Power Factor

Solution:

1. Capacitive Reactance = $X_C$ = $\frac{1}{2\pi f C}$

Since there is no capacitor in the circuit, C= 0

⇒$X_C = \frac{1}{0}$ = ∞

2. Inductive Reactance = $X_L = 2\pi f L$

= 100 X π X 0.02

= 2π

3. Impedance = $Z= \sqrt{R^2 + (X_L - X_C)^2}$

=$\sqrt{25 X 4 \pi^2}$ = $\sqrt{1000}$ = 10√10

4.Power Factor = cos∅ = R/Z = (5+5)/10√10

= 1/√10

Hence, the capacitive reactance, the inductive reactance, the power factor, and the impedance of the circuit are ∞, 2π, 1/√10, and 10√10 respectively.