# Calculate the resistance of the bulb when a 60watt light bulb operates at a voltage of 120V.

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**Hint:**As a very first step, one could read the question well and hence note down the given information. Then you could recall the expression for power in terms of voltage and resistance as per the requirement. You could also recall some other relation and then do required substitutions and hence find the answer.

**Complete step-by-step solution:**

In the question, we are given the power of a bulb to be 60watt and it is also said that it operates at 120V. We are supposed to calculate the resistance of the bulb using this given information.

In order to solve the given question, we have to recall a relation that relates power with voltage and resistance. You may be familiar with the relation for power in terms of current and resistance that is mathematically given by,

$P={{I}^{2}}R$…………………………….. (1)

Since, we have no idea of the current passing through the bulb, we could recall the Ohm’s law that is mathematically given by,

$I=\dfrac{V}{R}$…………………………… (2)

Now we could substitute the equation (2) into (1), to get, power in terms of voltage and resistance as,

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

On substituting the values,

$R=\dfrac{{{V}^{2}}}{P}=\dfrac{{{120}^{2}}}{60}$

$\therefore R=240\Omega $

Therefore, we found the resistance of the bulb to be $240\Omega $.

**Note:**You shouldn’t actually take this lengthy method. If you remember the direct relation, well and good, you could directly carry out the substitutions and hence find the answer. Since all the quantities are given in their respective SI units, one shouldn’t worry about anything but simple multiplication and division will do.

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