# Where to use $I^2R$ and $\frac{V^2}{R}$ for power [closed]

I was given the following question,

Which of the following statement(s) is/are correct?

(a) Resistance of the filament of the bulb is inversely proportional to the power of the bulb.

(b) Resistance of the filament of the bulb is directly proportional to the power of the bulb.

(c) Higher the wattage of the bulb, higher is the current that can be allowed to pass through a bulb.

(d) Higher the wattage of the bulb, lower is the current that can be allowed to pass through a bulb.

The answer is a and c, but why can't b and d be true? Using $I^2R$ formula.

Electric circuits are usually supplied by voltage sources. These maintain constant voltage. If you increase the resistance, the voltage source will decrease the current to keep the voltage unchanged (U = R * I). This leads to drop of power.

However, if you use a current source to supply the circuit, it would be different. The current source is maintaining the current constant. If you increase the resistance, the current source will increase the voltage to keep the current unchanged. This will increase the power.

Conclusion: the use of the 2 formulas depends on the type of source you connect to the circuit. Please note that voltage sources are used most of the time.

• @user154981 I find this a very good answer. – Wrichik Basu May 6 '17 at 2:19

For these formulas to work, only one term should vary and the rest must remain constant.†

For resistors in series, $I$ is is constant, so you can easily compare the power dissipation of the resistors using the formula $P=I^2R$, Since $I$ is constant, $P\propto R$.

For resistors in parallel, $V$ is constant, so you can easily compare the power dissipation of the resistors using the formula $P=\frac{V^2}{R}$. Since $V$ is constant, $P\propto \frac{1}{R}$.

In your question what is the parameter that is constant? It is the battery's voltage that the bulb is connected to. So current actually depends on the value of $R$ i.e, it is not constant, current will give different values for different R but V will be constant in each case. By replacing $I$ by by $\frac{V}{R}$, you will get the relation, $P\propto \frac{1}{R}$.

• Comments are not for extended discussion; this conversation has been moved to chat. – Manishearth May 5 '17 at 14:43

You are getting confused by the meaning of wattage of equipments. Specifically wattage or power is usually given under the assumption of constant voltage. Now you should be able to see which formula is to be used.

• V square/r is right.I think so – user154981 May 5 '17 at 10:34
• And by using that my answer comes – user154981 May 5 '17 at 10:35
• As i said , wattage of bulbs etc are usually given assuming voltage is fixed. As such V^2/r is the correct rule in this case. If you think my answer helped you understand , please select it as correct – ssj3892414 May 5 '17 at 10:40
• I upvoted ur answer and thanks for giving me a confidence that I can also be intelligent like you guys. – user154981 May 5 '17 at 11:20