# $P=V^2/R$ confusion

I am confused on how to apply the formula

$P=\frac{V^2}{R}$

If I am given a bulb say with power 60W and it is connected to a supply of 120V. Then the resistance of the bulb is 240$\Omega$ but if the bulb is connected to a resistor say 10 $\Omega$ in series then will the value of V to be put in the formula still be 120V ? And hence the resistance of bulb remain same? Or do we have to apply V'=V-I×10 (where I is the current in the circuit)to get net potential across the bulb ?

In a nutshell what is the V stand for in the formula - The supply voltage or the the potential drop across the bulb and why ?

P.S- This question came to my mind while solving this Question

• In the equation $P = V^2/R$ all of the values are for a single element. In other words, $V$ is the voltage dropped across the bulb, $R$ is the resistance of the bulb, and $P$ is the power dissipated by the bulb. If you add another element in series you've got to figure out the voltage dropped across each element individually, or use $P = I^2 R$ which may be a bit easier if the elements are in series. – DanielSank Feb 14 '16 at 7:47
• Even i thought it should be like that but my book says that the resistance in this case would be 120^2/60 i.e 240 ohms. It might be because changing the value of V would change the resistance of the bulb which should not happen also. – Jai Mahajan Feb 14 '16 at 7:53
• The voltage V in the formula is the potential difference across the resistor in which you desire to know the dissipated power. As simple as that :D – brainst Feb 14 '16 at 8:27

## 1 Answer

If you increase the resistance of the bulb by 10 ohm, the total resistance becomes 240 + 10 = 250 ohm. The voltage across this combined resistance remains the same(120 v). So, you just need to plug this new resistance value in the formula and that will give you the power being consumed by the combined resistance which gives us 57.6 Watt. As we can see that in this case the value of v is the voltage supply to the combined resistance in the circuit.