Electrical Circuits
Work and Power:
Calculating these two are actually really simple and straight forward. It's the same exact formula that you used in physical science class which most had last year. W=F*D
Calculating these two are actually really simple and straight forward. It's the same exact formula that you used in physical science class which most had last year. W=F*D
Work: Abbreviation: Unit: Symbol:
W Joules J
W Joules J
W=F*D
-The equation for work is the amount of force used in newtons multiplied by the amount of distance the object has moved in meters.
Example Problem:
A box is pushed a distance of 20 meters with an applied force of 8 Newtons in a time period of 30 seconds. How much work was put into pushing the box?
All you have to do is plug in the necessary units to find work. The time of 30 seconds is unnecessary, since force and distance are the only things required for calculating work. The distance pushed is 20 meters, which is one of the necessary units. The force applied is 8 Newtons.
Work: W= F*d= 8*20= 160. W=160 Joules required.
Example Problem:
A box is pushed a distance of 20 meters with an applied force of 8 Newtons in a time period of 30 seconds. How much work was put into pushing the box?
All you have to do is plug in the necessary units to find work. The time of 30 seconds is unnecessary, since force and distance are the only things required for calculating work. The distance pushed is 20 meters, which is one of the necessary units. The force applied is 8 Newtons.
Work: W= F*d= 8*20= 160. W=160 Joules required.
Power: Abbreviation: Unit: Symbol:
P Watts W
P=V*I
The equation for power is a little different in engineering than in physical science. The equation in engineering is mostly used for calculating something has to do with electricity. So, in this case, the equation is the current of the system times the voltage of the system.
Example Problem:
If the lighting in a house ran off a battery unit that produced 30 volts and had a current of 6, how much power would be produced?
Plug in the units units. The voltage is 30 and the current is 6.
Work: P=V*I=30*6=180. P=180 watts of power required.
Example Problem:
If the lighting in a house ran off a battery unit that produced 30 volts and had a current of 6, how much power would be produced?
Plug in the units units. The voltage is 30 and the current is 6.
Work: P=V*I=30*6=180. P=180 watts of power required.
Ohm's Law
-Mathematical relationship between the current, resistance, and voltage. It is three math formulas that can calculate the missing component (i.e. current, resistance, or voltage) in a circuit.
Abbreviation/Unit/Symbol
Voltage: V/Volts/V
Current: I/Amperes/A
Resistance: R/Ohms/Ω
Equations:
Problem #1: A flashlight has 150 Ohms of resistance and has a current of .04 Amperes. Find the voltage.
To find the voltage of this problem, think of the voltage, resistance, and current all on a pyramid with the voltage being on top and the current and resistance splitting the bottom. Cover up the voltage and you have only resistance and current left across from each other. To get the voltage, you would multiple these two numbers together to get the voltage.
Work: I*A= .04*150= 6 Volts
The voltage of the flashlight is 6.
Problem #2: A series circuit with one battery has a voltage of 9 and a resistance of 180. Find the current.
Go back to the pyramid and now cover up the current. All we have left is the voltage over the resistance. It should look like a division problem and that is exactly what to do. You divide the voltage by the resistance and that would equal the current.
Work: V/R= 9/180= .05 Amps
The current of this circuit is .05 Amps
Problem #3: A string of Christmas lights has a voltage of 18 and a current of .06. Find the resistance.
Once again, we go back to the pyramid. We cover up the resistance and see that only current and voltage are left. Just like before with resistance and voltage, you take the voltage divided by the current to get the resistance.
Work: V/I= 18/.06= 300 Ohms
The resistance of this strand is 300 Ohms
Problem #1: A flashlight has 150 Ohms of resistance and has a current of .04 Amperes. Find the voltage.
To find the voltage of this problem, think of the voltage, resistance, and current all on a pyramid with the voltage being on top and the current and resistance splitting the bottom. Cover up the voltage and you have only resistance and current left across from each other. To get the voltage, you would multiple these two numbers together to get the voltage.
Work: I*A= .04*150= 6 Volts
The voltage of the flashlight is 6.
Problem #2: A series circuit with one battery has a voltage of 9 and a resistance of 180. Find the current.
Go back to the pyramid and now cover up the current. All we have left is the voltage over the resistance. It should look like a division problem and that is exactly what to do. You divide the voltage by the resistance and that would equal the current.
Work: V/R= 9/180= .05 Amps
The current of this circuit is .05 Amps
Problem #3: A string of Christmas lights has a voltage of 18 and a current of .06. Find the resistance.
Once again, we go back to the pyramid. We cover up the resistance and see that only current and voltage are left. Just like before with resistance and voltage, you take the voltage divided by the current to get the resistance.
Work: V/I= 18/.06= 300 Ohms
The resistance of this strand is 300 Ohms