Power Factor worksheet

1. Apparent power (S): 62.22 kVA

1. Find the apparent power S: divide the real power by the power factor

   $S$$=\frac{56}{0.9}$$=62.2222222222\text{\hspace{0.17em}kVA}$

2. Find the reactive power Q. The three powers form a right triangle, so Q is the leftover side

   $Q$$=\sqrt{{62.2222222222}^2-{56}^2}$$=27.1220378709\text{\hspace{0.17em}kVAR}$

3. Find the phase angle φ, how far the current lags behind the voltage. Its cosine is the power factor

   $\phi$$=\text{arccos}\left(0.9\right)$$=25.8419327632\text{\hspace{0.17em}°}$

2. Power factor: 0.848

1. Find the apparent power S. The three powers form a right triangle, so S is the long side

   $S$$=\sqrt{{32}^2+{20}^2}$$=37.7359245282\text{\hspace{0.17em}kVA}$

2. Find the power factor (PF): divide the real power P, which does the work, by the apparent power S

   $\mathrm{PF}$$=\frac{32}{37.7359245282}$$=0.847998304$

3. Find the phase angle φ, how far the current lags behind the voltage. Its cosine is the power factor

   $\phi$$=\text{arccos}\left(0.847998304\right)$$=32.0053832081\text{\hspace{0.17em}°}$

4. Size the capacitor (Qc) to raise the power factor to 1. It must supply the reactive power you no longer want from the grid: the angle now (φ₁) versus the target angle (φ₂ = arccos(1) = 0°)

   $\mathrm{Qc}$$=32\times \left(0.625-0\right)$$=20\text{\hspace{0.17em}kVAR}$

5. Find the apparent power after correction (S₂): the same real power divided by the target power factor

   $\mathrm{S₂}$$=\frac{32}{1}$$=32\text{\hspace{0.17em}kVA}$

3. Apparent power (S): 45 VA

1. Find the apparent power S: divide the real power by the power factor

   $S$$=\frac{45}{1}$$=45\text{\hspace{0.17em}VA}$

2. Find the reactive power Q. The three powers form a right triangle, so Q is the leftover side

   $Q$$=\sqrt{{45}^2-{45}^2}$$=0\text{\hspace{0.17em}VAR}$

3. Find the phase angle φ, how far the current lags behind the voltage. Its cosine is the power factor

   $\phi$$=\text{arccos}\left(1\right)$$=0\text{\hspace{0.17em}°}$

4. Power factor: 0.6

1. Find the power factor (PF): divide the real power P, which does the work, by the apparent power S

   $\mathrm{PF}$$=\frac{33}{55}$$=0.6$

2. Find the reactive power Q. The three powers form a right triangle, so Q is the leftover side

   $Q$$=\sqrt{{55}^2-{33}^2}$$=44\text{\hspace{0.17em}VAR}$

3. Find the phase angle φ, how far the current lags behind the voltage. Its cosine is the power factor

   $\phi$$=\text{arccos}\left(0.6\right)$$=53.1301023542\text{\hspace{0.17em}°}$

4. Size the capacitor (Qc) to raise the power factor to 0.9. It must supply the reactive power you no longer want from the grid: the angle now (φ₁) versus the target angle (φ₂ = arccos(0.9) = 25.8419°)

   $\mathrm{Qc}$$=33\times \left(1.33333333-0.4843221\right)$$=28.0173705404\text{\hspace{0.17em}VAR}$

5. Find the apparent power after correction (S₂): the same real power divided by the target power factor

   $\mathrm{S₂}$$=\frac{33}{0.9}$$=36.6666666667\text{\hspace{0.17em}VA}$

5. Real power (P): 49 W

1. Find the real power P: multiply the apparent power by the power factor

   $P$$=70\times 0.7$$=49\text{\hspace{0.17em}W}$

2. Find the reactive power Q. The three powers form a right triangle, so Q is the leftover side

   $Q$$=\sqrt{{70}^2-{49}^2}$$=49.9899989998\text{\hspace{0.17em}VAR}$

3. Find the phase angle φ, how far the current lags behind the voltage. Its cosine is the power factor

   $\phi$$=\text{arccos}\left(0.7\right)$$=45.5729959992\text{\hspace{0.17em}°}$

4. Size the capacitor (Qc) to raise the power factor to 0.95. It must supply the reactive power you no longer want from the grid: the angle now (φ₁) versus the target angle (φ₂ = arccos(0.95) = 18.1949°)

   $\mathrm{Qc}$$=49\times \left(1.02020406-0.32868411\right)$$=33.884477846\text{\hspace{0.17em}VAR}$

5. Find the apparent power after correction (S₂): the same real power divided by the target power factor

   $\mathrm{S₂}$$=\frac{49}{0.95}$$=51.5789473684\text{\hspace{0.17em}VA}$

6. Power factor: 0.7679

1. Find the apparent power S: multiply volts by amps, then divide by 1000 to get kVA

   $S$$=\frac{280\times 200}{\mathrm{1,000}}$$=56\text{\hspace{0.17em}kVA}$

2. Find the power factor (PF): divide the real power P, which does the work, by the apparent power S

   $\mathrm{PF}$$=\frac{43}{56}$$=0.7678571429$

3. Find the reactive power Q. The three powers form a right triangle, so Q is the leftover side

   $Q$$=\sqrt{{56}^2-{43}^2}$$=35.8747822293\text{\hspace{0.17em}kVAR}$

4. Find the phase angle φ, how far the current lags behind the voltage. Its cosine is the power factor

   $\phi$$=\text{arccos}\left(0.7678571429\right)$$=39.8381498056\text{\hspace{0.17em}°}$

5. Size the capacitor (Qc) to raise the power factor to 0.95. It must supply the reactive power you no longer want from the grid: the angle now (φ₁) versus the target angle (φ₂ = arccos(0.95) = 18.1949°)

   $\mathrm{Qc}$$=43\times \left(0.83429726-0.32868411\right)$$=21.7413657066\text{\hspace{0.17em}kVAR}$

6. Find the apparent power after correction (S₂): the same real power divided by the target power factor

   $\mathrm{S₂}$$=\frac{43}{0.95}$$=45.2631578947\text{\hspace{0.17em}kVA}$

7. Power factor: 0.9333

1. Find the power factor (PF): divide the real power P, which does the work, by the apparent power S

   $\mathrm{PF}$$=\frac{42}{45}$$=0.9333333333$

2. Find the reactive power Q. The three powers form a right triangle, so Q is the leftover side

   $Q$$=\sqrt{{45}^2-{42}^2}$$=16.1554944214\text{\hspace{0.17em}kVAR}$

3. Find the phase angle φ, how far the current lags behind the voltage. Its cosine is the power factor

   $\phi$$=\text{arccos}\left(0.9333333333\right)$$=21.0394697813\text{\hspace{0.17em}°}$

8. Power factor: 0.727

1. Find the apparent power S. The three powers form a right triangle, so S is the long side

   $S$$=\sqrt{{36}^2+{34}^2}$$=49.5176736126\text{\hspace{0.17em}VA}$

2. Find the power factor (PF): divide the real power P, which does the work, by the apparent power S

   $\mathrm{PF}$$=\frac{36}{49.5176736126}$$=0.7270131525$

3. Find the phase angle φ, how far the current lags behind the voltage. Its cosine is the power factor

   $\phi$$=\text{arccos}\left(0.7270131525\right)$$=43.3634229584\text{\hspace{0.17em}°}$

4. Size the capacitor (Qc) to raise the power factor to 0.8. It must supply the reactive power you no longer want from the grid: the angle now (φ₁) versus the target angle (φ₂ = arccos(0.8) = 36.8699°)

   $\mathrm{Qc}$$=36\times \left(0.94444444-0.75\right)$$=7\text{\hspace{0.17em}VAR}$

5. Find the apparent power after correction (S₂): the same real power divided by the target power factor

   $\mathrm{S₂}$$=\frac{36}{0.8}$$=45\text{\hspace{0.17em}VA}$

9. Real power (P): 42 kW

1. Find the real power P: multiply the apparent power by the power factor

   $P$$=70\times 0.6$$=42\text{\hspace{0.17em}kW}$

2. Find the reactive power Q. The three powers form a right triangle, so Q is the leftover side

   $Q$$=\sqrt{{70}^2-{42}^2}$$=56\text{\hspace{0.17em}kVAR}$

3. Find the phase angle φ, how far the current lags behind the voltage. Its cosine is the power factor

   $\phi$$=\text{arccos}\left(0.6\right)$$=53.1301023542\text{\hspace{0.17em}°}$

10. Power factor: 0.7649

1. Find the apparent power S. The three powers form a right triangle, so S is the long side

   $S$$=\sqrt{{38}^2+{32}^2}$$=49.6789693935\text{\hspace{0.17em}kVA}$

2. Find the power factor (PF): divide the real power P, which does the work, by the apparent power S

   $\mathrm{PF}$$=\frac{38}{49.6789693935}$$=0.7649111981$

3. Find the phase angle φ, how far the current lags behind the voltage. Its cosine is the power factor

   $\phi$$=\text{arccos}\left(0.7649111981\right)$$=40.1009075462\text{\hspace{0.17em}°}$

4. Size the capacitor (Qc) to raise the power factor to 1. It must supply the reactive power you no longer want from the grid: the angle now (φ₁) versus the target angle (φ₂ = arccos(1) = 0°)

   $\mathrm{Qc}$$=38\times \left(0.84210526-0\right)$$=32\text{\hspace{0.17em}kVAR}$

5. Find the apparent power after correction (S₂): the same real power divided by the target power factor

   $\mathrm{S₂}$$=\frac{38}{1}$$=38\text{\hspace{0.17em}kVA}$