Question

A single-phase motor takes apparent power of 1909 VA at a power factor of 0.866 lagging when connected to a 230 V, 50 Hz supply. Two similar capacitors are then connected in parallel with each other to form a capacitance bank. The capacitor bank is now connected in parallel with the motor to raise the power factor to 0.95. Calculate the capacitance of each capacitor.​

Answer 1

We can use the following formula to solve this problem:

Q = P / tan(theta)

where Q is the reactive power in VARs, P is the active power in watts, and theta is the angle between the active and reactive power.

First, let's find out the active power of the single-phase motor:

P = apparent power x power factor
P = 1909 VA x 0.866
P = 1655.274 W

Next, let's find the reactive power of the single-phase motor:

Q = P / tan(theta)
tan(theta) = sqrt(1 - power factor^2) = sqrt(1 - 0.866^2) = 0.5
theta = arctan(0.5) = 26.565 degrees

Q = P / tan(theta)
Q = 1655.274 W / tan(26.565)
Q = 970.7 VARs

The power factor of the motor after connecting the capacitor bank is 0.95, so the angle between the active and reactive power is:

theta = arccos(0.95) = 18.19 degrees

To raise the power factor from 0.866 to 0.95, the reactive power should be reduced to:

Q' = P / tan(theta)
Q' = 1655.274 W / tan(18.19 degrees)
Q' = 584.11 VARs

The capacitance required to produce the required reactive power can be calculated using:

C = Q' / (2 * pi * f * V^2)
where C is the capacitance in farads, f is the frequency in Hz, and V is the voltage in volts.

C = 584.11 VARs / (2 * pi * 50 Hz * (230 V)^2)
C = 1.645 microfarads

Since we have two capacitors, the capacitance of each capacitor should be:

C1 = C2 = C/2 = 0.823 microfarads.

Therefore, each capacitor should be 0.823 microfarads to raise the power factor of the single-phase motor to 0.95.