# $P=V^2/R$ 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. Feb 14, 2016 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. Feb 14, 2016 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 Feb 14, 2016 at 8:27