A certain load takes 10kW at 0.5pf lagging when connected to a 230V, 60Hz source. The supply

company charges a penalty when the power factor fall below 0.8. What size of capacitor must be used in
order to avoid penalty?

To calculate the size of the capacitor needed to avoid the penalty, we can use the formula:

\[ Q_c = \sqrt{P^2 \cdot \left( \frac{1}{\tan^2(\theta_s)}-1 \right)} \]

Where:
- \( Q_c \) is the reactive power provided by the capacitor in VARs,
- \( P \) is the active power of the load in watts,
- \( \theta_s \) is the angle of the power factor required.

Given:
- The active power \( P = 10 \, \text{kW} \)
- The power factor \( \text{pf} = 0.5 \) (lagging)
- The source voltage \( V = 230 \, \text{V} \)
- The frequency \( f = 60 \, \text{Hz} \)
- The power factor required \( \text{pf}_\text{req} = 0.8 \)

First, we need to calculate the angle \( \theta_s \):

\[ \theta_s = \arccos(\text{pf}_\text{req}) = \arccos(0.8) \]

Now, let's calculate the reactive power \( Q_c \):

\[ Q_c = \sqrt{P^2 \cdot \left( \frac{1}{\tan^2(\theta_s)}-1 \right)} \]

Finally, we can calculate \( Q_c \) using the given values and the calculated \( \theta_s \):

Step 1: Calculate \( \theta_s \)
\[ \theta_s = \arccos(0.8) = 0.6435 \, \text{rad} \]

Step 2: Calculate \( Q_c \)
\[ Q_c = \sqrt{10^2 \cdot \left( \frac{1}{\tan^2(0.6435)}-1 \right)} \]

Using the tangent function:
\[ Q_c = \sqrt{10^2 \cdot \left( \frac{1}{\left( \frac{\sin(0.6435)}{\cos(0.6435)} \right)^2}-1 \right)} \]

Simplifying:
\[ Q_c = \sqrt{10^2 \cdot \left( \frac{1}{\left( \frac{\sin(0.6435)}{\cos(0.6435)} \right)^2}-1 \right)} \]

Calculating:
\[ Q_c = \sqrt{10^2 \cdot \left( \frac{1}{\left( \frac{0.5950}{0.8030} \right)^2}-1 \right)} \]

\[ Q_c = \sqrt{10^2 \cdot \left( \frac{1}{0.7415}-1 \right)} \]

\[ Q_c = \sqrt{10^2 \cdot 1.3484} \]

\[ Q_c = \sqrt{134.84} \]

\[ Q_c = 11.61 \, \text{VARs} \]

Therefore, the size of the capacitor needed to avoid the penalty is approximately 11.61 VARs.

To calculate the size of the capacitor needed to avoid penalty, we first need to determine the reactive power and the existing power factor of the load.

1. Calculate the apparent power (S):
Apparent Power (S) = Power (P) / Power Factor (PF)
S = 10kW / 0.5
S = 20kVA

2. Calculate the reactive power (Q):
Q = √(S^2 - P^2)
Q = √(20kVA^2 - 10kW^2)
Q = √(400kVA^2 - 100kVA^2)
Q = √300kVA^2
Q = 547.7kVAR

3. Calculate the required reactive power (Qr):
Qr = S * √(1 - PF^2)
Qr = 20kVA * √(1 - (0.8)^2)
Qr = 20kVA * √(1 - 0.64)
Qr = 20kVA * √0.36
Qr = 20kVA * 0.6
Qr = 12kVAR

4. Calculate the required capacitance (C):
Qr = (2 * π * f * C * V^2) / 1000
12kVAR = (2 * π * 60Hz * C * (230V)^2) / 1000
12kVAR = (2 * 3.14 * 60Hz * C * 230V * 230V) / 1000
12kVAR = (2 * 3.14 * 60Hz * C * 230V^2) / 1000
12kVAR = (2 * 3.14 * 60Hz * C * 52900V^2) / 1000
12kVAR = (2 * 3.14 * 60Hz * C * 52900V^2) / 1000
12kVAR = (6.28 * 60Hz * C * 52900V^2) / 1000
12kVAR = (3768 * C * 52900V^2) / 1000
12 = (3768 * C * (230)^2) / 1000
(3768 * C * (230)^2) / 1000 = 12
(3768 * C * 52900) / 1000 = 12
(3768 * C * 52900) = 12000
C = 12000 / (3768 * 52900)
C ≈ 6.42 μF

Therefore, a capacitor of approximately 6.42 μF must be used in order to avoid penalties for a power factor falling below 0.8.