Question

A RLC series circuit consists of R = 75 Ω, L = 125 mH and C = 200 µF. The circuit is
excited by a sinusoidal source of value 115 V, 60 Hz. Determine the voltage across
various elements. Calculate the current, active power, reactive and apparent
power. Also, draw the phasor diagram.

Answer 1

Answer:

The Active power(P) = 0.566 W, reactive power (Q) = 0.354 var , and apparent power (S) = 10VA.

Explanation:

To calculate the voltage across the various elements in the RLC series circuit and determine the current, active power, reactive power, and apparent power, we can follow these steps:

Step 1: Calculate the impedance (Z) of the circuit:

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

$X _ L = 2 \pi fL$

$X_C = \frac{ 1}{(2\pi fC)}$

Given:

R = 75 Ω

L = 125 mH = 0.125 H

C = 200 µF = 0.0002 F

V = 115 V

f = 60 Hz

$X_L = 2\pi (60)(0.125)$

= 47.1 Ω

$X_C = \frac{ 1}{2π(60)(0.0002)}$

= 1326.8 Ω

$Z = \sqrt{ (75^2 + (47.1 - 1326.8)^2)}$

Z ≈ 1326.9 Ω

Step 2: Calculate the current (I):

$I = \frac{ V}{Z}$

$I = \frac{115}{1326.9 }$

≈ 0.0868 A

Step 3: Calculate the voltage across each element:

Voltage across resistor

(VR) = I × R = 0.0868 × 75

≈ 6.51 V

Voltage across inductor

(VL) = I × XL = 0.0868 × 47.1

≈ 4.08 V

Voltage across capacitor

(VC) = I × XC = 0.0868 × 1326.8

≈ 115.24 V

Step 4: Calculate the active power (P), reactive power (Q), and apparent power (S):

P = VR × I = 6.51 × 0.0868 ≈ 0.566 W

Q = VL × I = 4.08 × 0.0868 ≈ 0.354 var (reactive power)

S = V × I = 115 × 0.0868 ≈ 10 VA (apparent power)

So, The Active power(P) = 0.566 W, reactive power (Q) = 0.354 var , and apparent power (S) = 10VA.

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