Watts to Amps worksheet

1. Current: 3.13 A

1. Start from how power works for single-phase AC: power (P) is voltage (V) times current (I) times the power factor (PF)

   $P$$=V\times I\times \mathrm{PF}$

2. To get the current on its own, divide the power by everything else it was multiplied by

   $I$$=\frac{P}{V\times \mathrm{PF}}$

3. Put in your numbers and work it out

   $I$$=\frac{\mathrm{1,500}\text{\hspace{0.17em}W}}{480\text{\hspace{0.17em}V}\times 1}$$=3.125\text{\hspace{0.17em}A}$

2. Power: 6,240 W

1. Start from how power works for DC: power (P) is voltage (V) times current (I)

   $P$$=V\times I$

2. Multiply the voltage by the current

   $P$$=208\text{\hspace{0.17em}V}\times 30\text{\hspace{0.17em}A}$$=\mathrm{6,240}\text{\hspace{0.17em}W}$

3. Power: 561.18 W

1. Start from how power works for three-phase AC: power (P) is the line-to-line voltage (V) times the current in each wire (I), times √3 and the power factor (PF)

   $P$$=V\times I\times \sqrt{3}\times \mathrm{PF}$

2. Multiply the voltage, the current, √3 and the power factor

   $P$$=24\text{\hspace{0.17em}V}\times 15\text{\hspace{0.17em}A}\times \sqrt{3}\times 0.9$$=561.184462\text{\hspace{0.17em}W}$

3. Find the apparent power, what the wiring really has to carry, by dividing the power by the power factor

   $S$$=\frac{561.184462}{0.9}$$=623.538291\text{\hspace{0.17em}VA}$

4. Current: 125 A

1. Start from how power works for DC: power (P) is voltage (V) times current (I)

   $P$$=V\times I$

2. To get the current on its own, divide the power by everything else it was multiplied by

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

3. Put in your numbers and work it out

   $I$$=\frac{\mathrm{1,500}\text{\hspace{0.17em}W}}{12\text{\hspace{0.17em}V}}$$=125\text{\hspace{0.17em}A}$

5. Voltage: 625 V

1. Start from how power works for single-phase AC: power (P) is voltage (V) times current (I) times the power factor (PF)

   $P$$=V\times I\times \mathrm{PF}$

2. To get the voltage on its own, divide the power by everything else it was multiplied by

   $V$$=\frac{P}{I\times \mathrm{PF}}$

3. Put in your numbers and work it out

   $V$$=\frac{\mathrm{1,000}\text{\hspace{0.17em}W}}{2\text{\hspace{0.17em}A}\times 0.8}$$=625\text{\hspace{0.17em}V}$

4. Find the apparent power, what the wiring really has to carry, by dividing the power by the power factor

   $S$$=\frac{\mathrm{1,000}}{0.8}$$=\mathrm{1,250}\text{\hspace{0.17em}VA}$

6. Power: 1,632 W

1. Start from how power works for single-phase AC: power (P) is voltage (V) times current (I) times the power factor (PF)

   $P$$=V\times I\times \mathrm{PF}$

2. Multiply the voltage, the current and the power factor

   $P$$=48\text{\hspace{0.17em}V}\times 40\text{\hspace{0.17em}A}\times 0.85$$=\mathrm{1,632}\text{\hspace{0.17em}W}$

3. Find the apparent power, what the wiring really has to carry, by dividing the power by the power factor

   $S$$=\frac{\mathrm{1,632}}{0.85}$$=\mathrm{1,920}\text{\hspace{0.17em}VA}$

7. Power: 684 W

1. Start from how power works for single-phase AC: power (P) is voltage (V) times current (I) times the power factor (PF)

   $P$$=V\times I\times \mathrm{PF}$

2. Multiply the voltage, the current and the power factor

   $P$$=24\text{\hspace{0.17em}V}\times 30\text{\hspace{0.17em}A}\times 0.95$$=684\text{\hspace{0.17em}W}$

3. Find the apparent power, what the wiring really has to carry, by dividing the power by the power factor

   $S$$=\frac{684}{0.95}$$=720\text{\hspace{0.17em}VA}$

8. Voltage: 1,052.63 V

1. Start from how power works for single-phase AC: power (P) is voltage (V) times current (I) times the power factor (PF)

   $P$$=V\times I\times \mathrm{PF}$

2. To get the voltage on its own, divide the power by everything else it was multiplied by

   $V$$=\frac{P}{I\times \mathrm{PF}}$

3. Put in your numbers and work it out

   $V$$=\frac{\mathrm{5,000}\text{\hspace{0.17em}W}}{5\text{\hspace{0.17em}A}\times 0.95}$$=\mathrm{1,052.631579}\text{\hspace{0.17em}V}$

4. Find the apparent power, what the wiring really has to carry, by dividing the power by the power factor

   $S$$=\frac{\mathrm{5,000}}{0.95}$$=\mathrm{5,263.157895}\text{\hspace{0.17em}VA}$

9. Current: 25 A

1. Start from how power works for DC: power (P) is voltage (V) times current (I)

   $P$$=V\times I$

2. To get the current on its own, divide the power by everything else it was multiplied by

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

3. Put in your numbers and work it out

   $I$$=\frac{\mathrm{1,200}\text{\hspace{0.17em}W}}{48\text{\hspace{0.17em}V}}$$=25\text{\hspace{0.17em}A}$

10. Voltage: 900 V

1. Start from how power works for DC: power (P) is voltage (V) times current (I)

   $P$$=V\times I$

2. To get the voltage on its own, divide the power by everything else it was multiplied by

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

3. Put in your numbers and work it out

   $V$$=\frac{\mathrm{1,800}\text{\hspace{0.17em}W}}{2\text{\hspace{0.17em}A}}$$=900\text{\hspace{0.17em}V}$