# Why can I use P = I²R but not P=V²/R when calculating energy lost in a circuit?

I am working through a book of problems and am confused by the answer to this one:

A 12 V battery supplies 60 A for 2 seconds.

The total resistance of the wires in the circuit is 0.01 Ohm.

Q1. What is the total power supplied?

Q2. What is the energy lost as heat in the wires?

A1:

Total power output = 12 * 60 * 2 = 1440 Joules.

All good so far.

A2:

This is the answer in the book:

P = I²R * t = 3600 * 0.01 * 2 = 72 Joules

That's fine with me. However, if I use the equivalent equation P=V²/R ...

P = V²/ R * t = 12² / 0.01 * 2 = 28,800 Joules

Both of these equations are for P, so how are they giving me different answers?

• There is more in the circuit than just wires...but they don't say what it is, so you can't make any calculations about it. – The Photon Dec 8 '16 at 5:48
• The V in the equation is not what you think it is. It's the voltage across the wire just like I is the current through the wire. – user253751 Dec 8 '16 at 20:46

60A through a 0.01ohm resistance gives a 600mV drop. That is the voltage you need to use in the equation.

• OK So I understand that generally speaking P=I^2R is used to find out power lost in resistance. When is P=V^2/R used? – thatsagoal Dec 8 '16 at 8:03
• @thatsagoal: When you have the voltage drop (V) and the load resistance (R). – Ignacio Vazquez-Abrams Dec 8 '16 at 8:15
• Use whichever one is easier any time you want. But the V has to be across whatever the I is through. If you don't know I, then that is a good time to use V^2/R. If you don't know V, then that is a good time to use I^2*R. – mkeith Dec 8 '16 at 8:21

The problem assumes you understand something that is not clearly spelled out: the wires and the (unknown) load are in series. Therefore they share the current, not the voltage of the battery.

That's the situation:

simulate this circuit – Schematic created using CircuitLab

As other have pointed out, the voltage drop across the wires is small given their small resistance.

What you know is that the same current flows both in the load and in the wires, hence that's the information you must use to calculate power lost in the wiring.

• Is it a convention to use the American symbol for the wire's resistance and the European one for load? – v7d8dpo4 Dec 9 '16 at 10:33
• @v7d8dpo4 Well, I just used the European resistor symbol as a quick way to depict an unknown box (so it is not implied it is a linear load, or a resistor, for what matters). Although I'm in Europe, I prefer the American symbol for resistors. The European one is too useful for "generic boxes", like impedances or non-linear (or generic) loads. Anyway the schematic was intended as a quickly understandable depiction of the physical situation, it has no pretense to be an industrial standard schematic. – Lorenzo Donati -- Codidact.com Dec 10 '16 at 9:38

This is the answer in the book: P = I²R * t = 3600 * 0.01 * 2 = 72 Joules

You need to get a better book then because that is plainly wrong. Power does equal I²R but it doesn't equal I²R * t. Energy = I²R * t.

What is the total power supplied?

The total load resistance (including wires) is 12V/60A = 0.2 ohms so the total power supplied is 144/0.2 = 720 watts

What is the energy lost as heat in the wires?

Power lost in the wires is 60² * 0.01 = 36 watts so energy delivered is this number multiplied by time (2 seconds) = 72 joules.

Why can I use P = I²R but not P=V²/R when calculating energy lost in a circuit?

Using ohms law, I = V/R therefore, I²R becomes (V/R)²R which becomes V²/R. Just make sure that the voltage you are talking about is across a resistor that has the current I flowing. Anything else is likely to be wrong or possibly "correct" by coincidence.

• Sorry I mixed up power and energy. I'll correct my question. – thatsagoal Dec 8 '16 at 11:26
• @thatsagoal No, don't do that because several people have pointed that out in their answers and you'll just end up p#s#i#g people off. I've rolled back the answer to the previous (warts and all). Please understand why I have done that. – Andy aka Dec 8 '16 at 11:28

If 12 V supplies 60 A, by Ohm's law the total resistance in the circuit must be $\frac{12~V}{60~A} = 0.2~\Omega$. Assuming the load resistor is in series with the wire, that means that the load resistor has a resistance of $0.2 - 0.01 = 0.19~\Omega$.

So, from Ohm's Law, the voltage drop across the load is $0.19 \cdot 60 = 11.4~V$ and the voltage drop across the wire is $0.01 \cdot 60 = 0.6~V$. This is the voltage that must be used with the wire resistance: $\frac{0.6^2}{0.01} \cdot 2 = 72~J$.

The misunderstanding in your original logic is that the wire resistance isn't the only resistance in the circuit.

• The OP accepted an answer in Dec 2016. Welcome to EE.SE. – Transistor Mar 27 '18 at 21:01

Power is energy per unit time. Energy is measured in joules, power in watts (joules/second).

The power lost in the wires is I^2*R.

Your energy calculation would be correct if the load of 190m$\Omega$ (12V/60A - 0.01$\Omega$) was replaced by a short circuit with just the wiring present. The current would be enormous (1200A) if the battery really held to 12V, and the wires would thus be dissipating 14.4kW and would quickly burn up. Not sure why you used 48V for the voltage though.

In the given case 95% of the energy makes it to the load, and 5% is lost in the wiring. The total power for the two seconds is 720W and 36W is lost in the wiring, leaving 684W for the load. In the two seconds, 72 joules heats the wiring.

First of all, You need to understand that all the power supplied by the battery is not equal to the power consumed by the wire. There will be losses within the battery itself.

for finding out power supplied by the battery we use the formula, P = V*I; where V is the voltage across its terminal and I is current flowing out positive terminal.

For finding out power consumed by the resistance of wire. we prefer, P = I^2*R; where I is the current flowing through the wire/resistance and R is the resistance offered by the wire.

We cannot use V^2/R because we are not sure about voltage across the resistance. There will always be resistance connected in series to any voltage source, which is resistance offered by the source itself. Voltage gets divided between internally offered resistance and a connected load. But current flowing through them is always same.

It is accurate to go with I square * R formula for finding out power consumed by resistance.