# Electricity and its units

1. Sep 4, 2015

### Elsa1234

According Ohm's laws V=IR , which means voltage is directly proportional to electricity, but according to the formula of P=VI= V=P/I, voltage is inversely proportional to electricity, so what's wrong here?

2. Sep 4, 2015

### e.bar.goum

You're comparing different things. And "electricity" doesn't come into V=IR , you mean "current". Ohms law tells you that the voltage is equal to current multiplied by resistance. So, if you increase the resistance, the current has to go down for constant voltage. V=P/I tells you that the voltage is equal to the power divided by current. So, if you increase the power, the current has to go up for constant voltage. Can you see that these are different things? Resistance and power aren't the same thing.

3. Sep 4, 2015

### Elsa1234

Oh Yeah sorry, I meant current :p , you're explaining well, can you elaborate please? See I got the V=IR formula but V=P/I means that for the same power an electric appliance consumes, if you increease the current the voltage would decrease or vice-versa, how?

Last edited: Sep 4, 2015
4. Sep 4, 2015

### e.bar.goum

I figured that's what you meant. I'd be happy to, but it might save us some time if you let me know what about the above explanation you didn't get.

In a physics sense, the Volt is equal to the potential difference between two parallel, infinite planes spaced 1 meter apart that create an electric field of 1 newton per coulomb.

That is,

$V = \frac{Potential Energy}{Charge}$

It always helps me to think about units explicitly. In SI base units, V = 1 V = 1 kg·m2·s−3·A−1. If you shuffle around the units of the equations you quoted, you'll find that they all give the same definition of V.

Or, by Ohms law and Joules law, we know that

$V = I * R = \frac{P}{I}$

In SI units,

$V = A * \Omega = \frac{W}{A}$

And if you juggle the units a bit, you quickly see that

$V = A * \Omega = \frac{W}{A} = \frac{J}{C}$

5. Sep 4, 2015

### Elsa1234

See I got the V=IR formula but V=P/I means that for the same power an electric appliance consumes, if you increease the current the voltage would decrease or vice-versa, how? I'm a grade 8 student, I don't clearly get the facts but I'm curious , so could you explain me by giving examples? :)

6. Sep 4, 2015

### Staff: Mentor

The two formulas are used for different things. Ohm's law shows a relationship between voltage, current, and resistance.
The other equation is used to show the relationship between power consumed, voltage, and current.

If I have a circuit and I measure a voltage of 10 volts and a current of 1 amp, then I know that the resistance of the circuit must be 10 ohms since R = V/I and 10/1 = 10.

That same circuit will be consuming 10 watts of power, which follows from the power equation P = IV: P = 1*10 = 10 watts

If I increase the voltage to 20 volts, then the circuit now has 2 amps running through it since: I = V/R, I = 20/10 = 2 amps

Since both the voltage and the current has increased, the new power consumed is: P = 2*20 = 40 watts.

Note that I never changed the resistance in the circuit, only the voltage. I could change the resistance, which will change the amount of current flow and the change in current flow will change the amount of power consumed, per the equations.

7. Sep 4, 2015

### Staff: Mentor

You should read $V=P/I$ to say that you can get the same amount of power out of the appliance if you double the voltage and halve the current. However, you have to remember that the current is still determined by the resistance ($V=IR$), so if you just double the voltage without changing the resistance, you'll get twice the current and twice the power (and your appliance will catch on fire because its wiring can't handle the increased current). Thus, if you're going to do something to change the voltage while keeping the power constant, you have to also do something to change the resistance so that the current changes too.

An example:
The table saw in my basement is powered by a 1.5 horsepower electric motor - that's the power output given by $P=IV$. Inside the motor there are connections that can be configured to change the resistance of the motor windings so that the motor produces that 1.5 HP using either 20 amperes and 120 volts or 10 amperes and 240 volts - convenient, because those are the two voltage levels widely available in North America.

Last edited: Sep 4, 2015
8. Sep 4, 2015

### Staff: Mentor

Use those two equations to find the relationship between P and R in terms of I and V.

9. Sep 4, 2015

### Staff: Mentor

For some devices you can hold power constant when varying voltage (motors), but for others (resistors) you can't, so you need to know which equation to apply in which situation. Generally, you can hold other things constant and vary the current.

10. Sep 4, 2015

### CWatters

+1 to the

As others have said. The short answer is that P isn't a constant. It's also dependant on the current. Your problem is actually nothing to do with Ohms law and electricity....

Take any old equation such as this one that I just made up...

A = B/C

At first glance it looks like A is inversely proportional to C.

But If I then told you that B = C2 you would get a different answer. By substituting you get..

A = C2/C = C

So A is directly proportional to C.