# Power for Electrical Contractor’s exam

Power is measured in watts (W) and is represented by the letter P The formula below shows Power designed in terms of voltage, current, and resistance. Along with Ohm’s law, their formulas can be a powerful tool when taking the Electrical Contractor’s exam. When solved for each of its variables, we get two extra forms of each of the power formulas.

Power (P) = Voltage (V) x Current (I)

Watts (W) = Volts (V) x Amperes (A)

Watts (W) = Volts (V) x Amperes (A)

Power (P) = Resistance (R) x Current (I)

Watts (W) = Ohms (Ω) x Amperes (A)

^{2}Watts (W) = Ohms (Ω) x Amperes (A)

^{2}Power (P) = Resistance (R) x Current (I)

Watts (W) = Ohms (Ω) x Amperes (A)

^{2}Watts (W) = Ohms (Ω) x Amperes (A)

^{2}Power (P) = Voltage (V)

Watts (W) = Volts (V)/Ohms (Ω)

^{2}/Resistance (R)Watts (W) = Volts (V)/Ohms (Ω)

P = V x I

I = P/V

V = P/I

I = P/V

V = P/I

P = R x I

I = P/I

V = √P/R

^{2}I = P/I

^{2}V = √P/R

P = V

R = V

V = √P x R

^{2}/RR = V

^{2}/PV = √P x R

## Example 1:

What is the power rating (in Watts) of an appliance that when connected to 120V pulls 30A of current?

To solve, use the formula which defines Power in terms of Voltage and current.

P = V x I

P = 120V x 30A

P = 3,600 W

To solve, use the formula which defines Power in terms of Voltage and current.

P = V x I

P = 120V x 30A

P = 3,600 W

## Example 2:

What's the current an appliance rating for 1,560 W will pull when connected to 120V?

To solve, use the formula which defines Current in terms of power and voltage.

I = P / V

I = 1,560W / 120V

I = 13A

To solve, use the formula which defines Current in terms of power and voltage.

I = P / V

I = 1,560W / 120V

I = 13A

## Example 3:

What Voltage does an appliance need to be connected to in order to pull 10A of current and have a power rating of 2,200W?

To solve, use the formula which defines voltage in terms of current and power.

V = P / I

V = 2,200W / 10A

V= 220V

To solve, use the formula which defines voltage in terms of current and power.

V = P / I

V = 2,200W / 10A

V= 220V

## Example 4:

What is the power when a resistance fo 18 Ω and a current of 10A are present?

P = R x I

P = 18 Ω x 10

P = 18 Ω x 100A

P = 1,800W

P = R x I

^{2}P = 18 Ω x 10

^{2}AP = 18 Ω x 100A

P = 1,800W

## Example 5:

What is the resistance of an appliance with a power rating of 2,800W and a 20A current?

R = P / I

R = 2,800W / 20

R = 2,800W / 20A

R = 7 Ω

R = P / I

^{2}R = 2,800W / 20

^{2}AR = 2,800W / 20A

R = 7 Ω

## Example 6:

What's the current pulled by an appliance with a power rating of 2,000W and a resistance of 20 Ω?

I = P/R

I = 2,000W / 20 Ω

I = 100A

I = 10A

I = P/R

I = 2,000W / 20 Ω

I = 100A

I = 10A

## Example 7:

What's the power (in Watts) of an appliance connected to 110V with a resistance of 30 Ω?

P = V

P = 110

P = 12,100V / 30 Ω

P = 403.33W

P = V

^{2}/ RP = 110

^{2}V / 30 ΩP = 12,100V / 30 Ω

P = 403.33W

## Example 8:

What's the resistance of an appliance connected to 210V with a power rating of 3,000W?

R = V

R = 210

R = 44,100V / 3,000W

R = 14.7 Ω

R = V

^{2}/ PR = 210

^{2}V / 3,000WR = 44,100V / 3,000W

R = 14.7 Ω

## Example 9:

What's the voltage when the power of 1,470W and a resistance of 30 Ω are present?

V = P x R

V = 1470W x 30 Ω

V = 44,100V

V = 210V

V = P x R

V = 1470W x 30 Ω

V = 44,100V

V = 210V

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