Key Takeaways
- Understanding Ohm's Law is crucial for determining the relationship between voltage, current, and resistance.
- Knowledge of power calculations allows for accurate assessments of energy transfer in electrical systems.
- Practical application of these calculations enables effective planning and troubleshooting in electrical work.
Ohm's Law
Understanding Voltage, Amperage, and Resistance
You are already familiar with Ohm's Law, the fundamental relationship between voltage (E), amperage (I), and resistance (R). This concept is the cornerstone of understanding electrical circuits, exclusively representing how these three elements interact without considering power or efficiency metrics which involve energy transfer connotated by joules.
|
Component |
Symbol |
Unit |
Description |
|
Voltage |
E |
Volts (V) |
The potential difference or electrical pressure |
|
Amperage |
I |
Amps (A) |
The flow rate of electric charge |
|
Resistance |
R |
Ohms (Ω) |
The opposition to the flow of an electric current |
Using Ohm's Law, adjustments to one variable affect the others, holding the relationship constant:
[ E = I \times R ]
Solving for Voltage (E=IR)
When you're looking to calculate voltage, you apply the formula:
[ E = I \times R ]
In practice, if you have a circuit with an amperage of 20A flowing through a 6Ω resistor, the voltage across that resistor is:
[ E = 20A \times 6Ω = 120V ]
Solving for Amperage (I=E/R)
To find amperage, the formula gets rearranged to:
[ I = \frac{E}{R} ]
For instance, with a 120V circuit across a 6Ω resistor, the amperage is:
[ I = \frac{120V}{6Ω} = 20A ]
Solving for Resistance (R=E/I)
Lastly, to discover the resistance, use the equation:
[ R = \frac{E}{I} ]
When you know the circuit draws 20A and has a voltage of 120V, the resistance calculates to:
[ R = \frac{120V}{20A} = 6Ω ]
This method of calculating using Ohm's Law is beneficial for quick, reliable results in practical electrical tasks.
Power Calculation
Defining Joules Law
Joules Law relates to the transfer of electrical power from one system to another, quantifying energy transfer in terms of joules. When electricity flows through an incandescent bulb, for instance, electrical energy is converted into light and heat. This energy transfer is time-dependent, yet amperage (current rate per second) inherently includes this temporal element. Therefore, for electrical power calculations, the time factor is already accounted for within the amperage component.
Calculating Power (P=IE)
To calculate power in watts, use the formula P = I × E, where P stands for power in watts, I for current in amperes, and E for voltage in volts. Think of it as Pi (π), which helps in memorizing the formula. If you have a circuit with 120 volts and 20 amperes, the total power is P = 20A × 120V = 2400W.
- ( P ) (Watts) = ( I ) (Amperes) × ( E ) (Volts)
- Example: ( P = 20A × 120V = 2400W )
Determining Amperage with Power
Amperage can be determined from power using the rearranged formula, ( I = \frac{P}{E} ). Given a 2400 watt circuit powered at 120 volts, the current would be ( I = \frac{2400W}{120V} = 20A ).
|
Power (P) |
Voltage (E) |
Current (I) |
|
2400W |
120V |
20A |
Determining Voltage with Power
To determine voltage when power and current values are known, the formula ( E = \frac{P}{I} ) can be used. For a 2400 watt circuit with a current of 20 amperes, the voltage would be ( E = \frac{2400W}{20A} = 120V ).
- ( E = \frac{P}{I} )
- Given: ( P = 2400W ), Current ( I = 20A )
- Voltage ( E = 120V )
|
Power (P) |
Current (I) |
Voltage (E) |
|
2400W |
20A |
120V |
Voltage Drop
Single Phase Voltage Drop Calculation
In a single phase circuit, the voltage drop can be determined through a specific formula. If you're dealing with two conductors, remember to use the factor of 2 which accounts for the return path. The length (L) of your circuit plays a crucial role, measured one way only as the formula incorporates the return trip. Use the constant K based on the material; for copper, it's 12.9. Your amperage (I), such as a 20-amp motor load, is plugged directly into the formula. On the denominator, input the circular mills (CM), which for a number 12 conductor is 6530.
Three Phase Voltage Drop Calculation
For three phase circuits, use the formula that involves the multiplier 1.732 (square root of three), indicative of the difference between single phase and three phase systems. This reflects in typically less voltage drop than single phase. As in single phase calculations, parameters like the material constant K for copper (12.9) are used, along with the circular mills for the conductor under consideration, for example, 6530 for a number 12.
Understanding the Use of Constants (K)
Constants such as K are essential for voltage drop calculations as they reflect the material properties, like the resistivity of copper (12.9) and aluminum (21.2). Memorize these constants for accurate computations in any scenario. Your choice of material will significantly impact the voltage drop due to its inherent resistance which is factored into the K constant.
Calculating with Circular Mills
The cross-sectional area of a conductor is central to determining resistance, and thus, voltage drop. This area is quantified in circular mills (CM). For example, a #12 conductor has a set value of 6530 circular mills. Larger conductors, such as 600 kcmil, are directly denoted by their area in circular mills, such as 600,000 circular mills. It's important to use the correct area in your calculations as it's directly proportional to the conductor's ability to carry current with minimum voltage drop.
Practical Application of Calculations
Ohm's Law is fundamental to your work as an electrician, providing essential insights into voltage (E), amperage (I), and resistance (R). Utilizing the easily remembered variations of the law is critical for quick calculations:
- When solving for voltage, use the equation E = I * R. For instance, if you have a current of 20 amps through a 6-ohm resistor, the voltage is 120 volts.
- To find amperage, I, the equation simplifies to I = E / R. With a known 120-volt circuit and a 6-ohm resistor, the current is 20 amps.
- For resistance, using the formula R = E / I is effective. Knowing a circuit draws 20 amps at 120 volts allows you to determine the resistance, which in this case is 6 ohms.
Moving on to Joule's Law — a concept you should equate with the energy transfer within a system. Here, you'll predominantly use the P = I * E equation, remembering it as “PIE” for power calculations.
With a 120-volt circuit drawing 20 amps:
- The total power can be determined by P = I * E, which equals 2400 watts.
- To find the current that a circuit draws, divide the power by the voltage, I = P / E, yielding 20 amps for a circuit with 2400 watts and 120 volts.
- To calculate the total wattage on a 20-amp circuit at 120 volts, simply multiply the two: P = I * E, resulting in 2400 watts.
Voltage drop becomes a critical calculation in the field. For single-phase circuits, the formula VD = (2 * K * I * L) / CM simplifies the process:
- L denotes the one-way length in feet you're dealing with in the field. For a 200-foot circuit, you'd use 100 feet for L due to the return path.
- K stands for the constant representing the material's resistivity. Use 12.9 for copper and 21.2 for aluminum.
- I is the load current in amps, which might be, for example, 20 amps for a specific motor.
- CM is the conductor's size in circular mills. For number 12 wire, use 6530 circular mills.
For a single-phase circuit using copper with a load current of 20 amps over a distance of 100 feet, you'd compute the voltage drop by plugging these values into your formula. Determining the correct voltage drop is crucial, especially when working with longer distances, to ensure proper voltage at the point of use and adequate performance of electrical equipment.
Remember to refer to chapter nine, table eight in the National Electric Code (NEC) for specific conductor sizes when calculating voltage drop, making sure to stay within acceptable limits for efficient and safe electrical system operation.