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Alex73 [517]
3 years ago
14

I NEED HELP PLEASE, THANKS! :)

Physics
2 answers:
brilliants [131]3 years ago
8 0

Answer:

P = VI = (IR)I = I2R

Explanation:

What the equation means is that if you double the current you end up with 4 times the power loss. It's like the area of carpet you need for a room - if you make the room twice as long and twice as wide you need 4x as much carpet. The physical explanation is that the voltage difference along a wire depends on the current - more current flowing with a resistance means more voltage (pressure of electricity if you like) is built up.

This extra voltage means more power. So if you double the current your would double the power, but you also double the voltage which doubles the power again = 4x as much power. P = VI = (IR)I = I2R

I hope this helps you out, if I'm wrong, just tell me.

Mumz [18]3 years ago
3 0

As I mentioned earlier, Ohm's law gives us the formula P = IV, where V is the voltage ( also known as the electrical potential difference ) and I is the current. It is confusing that P = I²R and P = IV are one in the same - so I want to go a bit deeper on that.

We have three formulas, P = IV, P = I²R, and P = V² / R. Each are considered the same. The two formulas P = I²R, and P = V² / R are derived from the statement that P = IV, under the condition V = IR. Substitute the value of V from this second condition V = IR into P = IV. You would get the following -

P = I( IR ),

P = I²R

That is how one can derive the formula P = I²R, and how P = IV and P = I²R are thought to be one in the same. If you would like, take a look at how to get the formula " P = V² / R, "

V = IR, P = IV

I = V / R, P = IV

P = ( V / R )V,

P = V² / R

<u><em>Hope that helps!</em></u>

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<em>In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law the current through the circuit is _____________ Amps. </em>

According to Ohm’s law V =I*R  

Where V = Voltage, I = Current and R = Resistance

I = V/R =6/3 =2 Amps of current

Answer = 2 Amps

<em> If a second identical lamp is connected in series, the 6-volt battery must push a charge through a total resistance of ________ Ω</em>.  <em>The current in the circuit is then __________ Amps. </em>

Since the second lamp is connected in series with the first, the total resistance will R₂ = 3 + 3 = 6Ω

2 resistance in series  R₂ =  6Ω

The current in the circuit with the two lamps connected in series is I₂ =V/R₂ =6/6 = 1 Amps

The current is 1 Amps

Answer =  6Ω  and 1 Amps

<em>If a third identical lamp is connected in series, the total resistance is now _________Ω.  The current through all three lamps in series is now _________ Amps.  </em>

Since the third lamp is connected in series with the first and second, the total resistance will R₃ = 3 + 3 + 3 = 9Ω

total resistance of the 3 lamps R₃ =  9Ω

The current in the circuit with the three lamps connected in series is

I =V/R₃ =6/9 = 0.67 Amps

The current through the 3 lamps I₃ = 0.67 Amps

Answer =  9Ω  and 0.67 Amps

<em>The current through each individual lamp is __________ Amps.  </em>

Since all 3 lamps are connected in series, the same current will flow through each of the  3 lamps, and that current is I₃  

The current through each individual lamp is 0.67 Amp

Answer = 0.67 Amp

<em>What is the power when a voltage of 120 volts drives a current of 3 amps through a device?  </em>

The formula for power P = I*V =120*3 = 360 Watts

power P  = 360 Watts

Answer = 360 Watts

<em>What is the current when a 90-W light bulb is connected to 120 V? </em>

From P =I*V, make I the subject of the formula, I = P/V =90/120 = 0.75

Current= 0.75 Amps

Answer =  0.75 Amps

<em>How much current does a 75-W light bulb draw when connected to 120 V?</em>  

Current I =P/V = 75/120 = 0.625 Amps.

Answer = 0.625 Amps

<em>If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it? </em>

Voltage V =P/I =6/3 =2 Volts

Answer = 2 Volts

<em> If a 60 W light bulb at 120 V is left on in your house to prevent burglary, and the power company charges 10 cents per kilowatt-hour, how much will it cost to leave the bulb on for 30 days? Show your work. </em>

24 hours make 1 day, so the number of hours the bulb was left on = 24 *30 = 720 hours

The power rating of the bulb is 60W = 60/1000 = 0.06 KiloWatt

Total power consumed in kilowatt-hour = 0.06 * 720 = 43.2 kilowatt-hour

Cost for 30 days = 0.1*43.2 = $4.32 ( note that 10 Cents = $0.1)

Answer =  $4.32

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