So our topic now is on electrical power. The formulas are quite simple but I feel like we weren't given everything we need.
We have
P = VI = (I^2)(R)=V^2/R
where P = Power
V = Volts
I = current
R = resistance
the questions are
1. an Electric furnace operating at 120V requires 3.0horse Power. Calculate the current and resistance. Where does horse power come into play? How do I use that in the formula? Can anyone please provide the formula which includes horsepower? :\ it says the answer is 18Amperes and 6.7 Ohms
2. A light bulb whose power is 100W when it is operated at 240V is instead connected to a 120V source. What is the current in the bulb? (Answer is 0.21A) How much power does it dissipate? (Answer is 25W)
though I know the formula here, having two Volt sources is confusing me. Also, where id 0.21 come from? And what does it mean by how much power does it dissipate?
so these are practice problems but I just don't know what formulas to use to get the answer....
Answers & Comments
Verified answer
ques 1.
One horsepower = 746 watts (approx, closer to 745.7 watts)
P = VI
I = P/V
= (3 x 746)120
= 18.65 amps
P = V^2/R
R = V^2/P
= 14400/(746 x 3)
= 6.434 ohms
ques 2.
current, voltage and power are values that can be changed. no matter how you change those three, the resistance stays constant.
so find the lightbulb's resistance first
P = V^2/R
R = V^2/P
= 240^2/100
= 576 ohms
now use that value for R when you change voltage from 240 volts to 120 volts
V = IR
I = V/R
= 120/576
= .208 amps
P = V^2/R
= 120^2/576
= 25 watts
you could also have found the new power dissipation by using the new found value of I
eg.
P = VI
= 120 x .208333333333 = 24.999999999 = 25 watts
Power dissipation is just another way of saying how many watts something will use at a given voltage. In a simple resistive circuit, you apply a voltage (volts) to a resistor (light bulb). The resistor draws current (amps), and uses power (watts). If you change the voltage, the resistor, which stays the same value, will draw a different amount of current and use a different amount of power.
Everyone who has ever started to learn basic electrical circuits has always come across these questions where you need to find an unknown value first before you can move on and find the answer to the actual question posed. For electrical circuits it as always good practice to make a list of the given or known values, like so
P = P2 =
V = V2 =
I = I2 =
R = R = etc.......
and fill in the values that are given. It is a lot easier then to see the empty spots are what you will most likely need to find mathematically.
ques 1.
One horsepower = 746 watts (approx, closer to 745.7 watts)
P = VI
I = P/V
= (3 x 746)120
= 18.65 amps
P = V^2/R
R = V^2/P
= 14400/(746 x 3)
= 6.434 ohms
ques 2.
current, voltage and power are values that can be changed. no matter how you change those three, the resistance stays constant.
so find the lightbulb's resistance first
P = V^2/R
R = V^2/P
= 240^2/100
= 576 ohms
now use that value for R when you change voltage from 240 volts to 120 volts
V = IR
I = V/R
= 120/576
= .208 amps
P = V^2/R
= 120^2/576
= 25 watts
you could also have found the new power dissipation by using the new found value of I
eg.
P = VI
= 120 x .208333333333 = 24.999999999 = 25 watts
Power dissipation is just another way of saying how many watts something will use at a given voltage. In a simple resistive circuit, you apply a voltage (volts) to a resistor (light bulb). The resistor draws current (amps), and uses power (watts). If you change the voltage, the resistor, which stays the same value, will draw a different amount of current and use a different amount of power.
Everyone who has ever started to learn basic electrical circuits has always come across these questions where you need to find an unknown value first before you can move on and find the answer to the actual question posed. For electrical circuits it as always good practice to make a list of the given or known values, like so
P = P2 =
V = V2 =
I = I2 =
R = R = etc.......
and fill in the values that are given. It is a lot easier then to see the empty spots are what you will most likely need to find mathematically.
Both of these problems are very poor.
The first because horsepower is not an SI unit and the conversions are unwise.
For a start a 1 hp motor is supposed to PRODUCE 1 hp but it needs more input power than this because of losses.
It would have been far better to rate the input power in kw.
The second is a poor question because if you do the experiment you find that the answers are wrong.
A light bulb changes its resistance with temperature. As do most resistors.
Their resistance is not a constant.
Now back to the first question. Multiple sources quote the conversion as 1 hp = 0.745699872 kw
so your motor requires 3.0 * 0.746 = 2.24 kw
power = VI sp I = power / V = 2240 /120 = 19 A
and power = v^2 / r
so r = v^2/p = 120^2/2240 = 6.43 ohm
For the second one you firstly find the resistance that would give 100 W with 240 V
( 100 = 240^2 / R)
then assume that this never changes, and use it with any supply volts to see what would happen.
1.
A horse-power (1 hp) is a unit of power. You are meant to look-up how to convert it to watts.
for electrical horse-power: 1hp = 746W
for mechanical or hydraulic horse-power: 1hp = 745.7W
We use the electrical value (though it won't make much difference to the result).
3hp = 3 x746 = = 2238W
P = Vi so
2238 = 120i
i = 2238/120 = 18.65A = 19A to 2 significant figures (not 18A)
R = V/i = 120/18.65 = 6.4Ω (not 6.7Ω)
I'm pretty sure my values are correct and your book's are wrong!
______________________________
2.
There are different ways to do this. One way it to start by working out the bulb's resistance. First when the bulb is connected to a 240V supply:
P = V²/R
R = V²/P = 240²/100 = 576Ω
When the bulb is connected to a 120V supply. :
i = V/R = 120/576 = 0.21A
(Note we assume R doesn't change. This is technically not correct, as the bulb will operate at a lower temperature when run at 120V; this reduced the resistance slightly. However this is a bit advanced and you can ignore it in this problem.)
Now work out the power when the bulb is connected to 120V:
P = V²/R = 120²/576 = 25W
(If you are good at maths,you can spot the P is optional to V² (assuming R is constant). Halving voltage, reduces power by 2² =4 times. So the new power = 100/4 = 25W.)
When it says "how much power does it dissipate?" it means how much heat and light energy is produced each second. For example if the bulb dissipates 100W, that means it turns 100J of electrical energy to 100J of heat and light energy each second.