Friday, October 18, 2019

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Books & Classes for your Contractor's License

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)

Power (P) = Resistance (R) x Current (I)2

Watts (W) = Ohms (Ω) x Amperes (A)2

Power (P) = Resistance (R) x Current (I)2

Watts (W) = Ohms (Ω) x Amperes (A)2

Power (P) = Voltage (V)2/Resistance (R)

Watts (W) = Volts (V)/Ohms (Ω)

P = V x I

I = P/V

V = P/I

P = R x I2

I = P/I2

V = √P/R

P = V2/R

R = V2/P

V = √P x R

Example 1: What is the power rating (in Watts) of an appliance that when connected to 120V pulls 15A of current? To solve, use the formula which defines Power in terms of Voltage and current.

P = V x I

P = 120V x 15A

P = 1,800 W

Example 2: What's the current an appliance rating for 2,400 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 = 2,400W / 120V

I = 20A

Example 3: What Voltage does an appliance need to be connected to in order to pull 30A of current and have a power rating of 6,600W? To solve, use the formula which defines voltage in terms of current and power.

V = P / I

V = 6,600W / 30A

V= 220V

Example 4: What is the power when a resistance fo 12 Ω and a current of 15A are present?

P = R x I2

P = 12 Ω x 152 A

P = 12 Ω x 225A

P = 2,700W

Example 5: What is the resistance of an appliance with a power rating of 3000W and a 10A current?

R = P / I2

R = 3,000W / 102 A

R = 3,000W / 100A

R = 30 Ω

Example 6: What's the current pulled by an appliance with a power rating of 1,000W and a resistance of 10 Ω?

I = P/R

I = 1,000W/10 Ω

I = 100A

I = 10A

Example 7: What's the power (in Watts) of an appliance connected to 240V with a resistance of 100 Ω?

P = V2 / R

P = 2402 V / 100 Ω

P = 57,600V / 100 Ω

P = 576W

Example 8: What's the resistance of an appliance connected to 220V with a power rating of 3,400W?

R = V2 / P

R = 2202 V / 3,400W

R = 48,400V / 3,400W

R = 14.23 Ω

Example 9: What's the voltage when the power of 960W and a resistance of 60 Ω are present?

V = P x R

V = 960W x 60 Ω

V = 57,600V

V = 240V

Ohms Law For Contractors Exam

Ohms law can be thought of as the relationship between voltage, current, and resistance

Voltage is measured in Volts (V) and is represented by the letter V

Current is measured in Amperes (A) and is represented by the letter I

Resistance is measured in Ohms (Ω) and is represented by the letter R

Current (I) = Voltage (V) / Resistance (R)

Amperes (A) = Volts (V) / Ohms (Ω)

Voltage (V) = Current (I) x Resistance (R)

Volts (V) = Amperes (A) x Ohms (Ω)

Example 1:

What's the voltage needed to generate 15 Amperes of Current in 8 ohms of Resistance?

To solve, use the formula for voltage.

V = I x R

V = 15A x 8 Ω

V = 120V

Example 2:

What is the current generated by a voltage of 120V through a resistance of 15 ohms?

To solve, use the formula for current.

I = V/R

I = 120V/15 Ω

I = 8A

Example 3:

What's the Resistance present if a voltage of 120V generates 3 Amperes of current?

To solve, use the formula for resistance.

R = V/I

R = 120V/3A

R = 40 Ω

Number of Shingle Courses for Roofing Contractor License

To calculate the number of shingle course on a roof, follow the formula below:

Number of Courses = Width of Roof Section (eave to ridge along common rafter)/Exposure

Example: Calculate the number of courses of 12-inch-wide asphalt shingles required to cover the roof below. Assume a slope of 6 in 12 and 5-inch exposure.


On the roof plan, the distance from the eave to the ridge is 20’. We use the table below to identify the actual distance


  • The common/jack factor for 6 in 12 slope is 1.118.
  • 20’ x 1.118 = 22.36' actual length
  • Number of courses = 22.36' / 5"

Note: before dividing, we must standardize our units. They must be both inches or feet. Let’s multiply the numerator (22.36’) times 12 to convert it to inches.

22.36 x 12 / 5 = 53.66 or 54 Courses.

This means that one of the two sections of the roof has 54 courses since the other section is exactly the same, we simply multiply times 2 to get the total number of courses. 54 x 2 = 108

General Contractor Experience

You must have at least one year of experience in the construction of structures at least four stories in height.

Applicants for the Certified General Contractor License must have experience in four or more of the experience areas listed below:

  • Foundation/Slabs greater than 20,000 sqft.
  • Masonry walls
  • Steel erection
  • Precast concrete structures
  • Column erection
  • Formwork for structural reinforced concrete
  • Elevated slabs