How much charge runs through a $ 100{\text{ W}} $ light bulb connected to a $ 120{\text{ V}} $ source for one hour?

Hint :Here, as power and Voltage is given so we will be using the expression $ {\text{P = V}} \times {\text{I}} $ to find out the current first and in order to calculate the charge running through the bulb, we will use $ {\text{q = i}} \times {\text{t}} $ later.

Complete Step By Step Answer:
We know that the electric power in terms of voltage and current is given by,
 $ {\text{P = V}} \times {\text{I}} $
Where, P is power in wattage
V is Voltage in Volt
And, I is current in Ampere.
As given in question we have Power, $ {\text{P = 100 W}} $ and Voltage, $ {\text{V = 120 V}} $ , Putting these values in above equation we will have,
 $ {\text{100 = 120}} \times {\text{I}} $
 $ {\text{I = 0}}{\text{.83 A}} $
So, we have got current $ {\text{I = 0}}{\text{.83 A}} $ and as asked in question we will now find out the charge.
For charge, we will be using, $ q = i \times t $
Where, q is charged in Coulomb, I is current in ampere and t is time in seconds for which current was/is running.
Now as we have time t in an hour we will convert it into seconds.
 $ 1{\text{ h = 60}} \times {\text{60 seconds}} $
Now, putting the value of I and t in equation above, we will have,
 $ q = 0.83 \times 3600 $
 $ q = 2988{\text{ C}} $
Hence, the answer is $ 2988 $ Coulomb charge will be running through the bulb in one hour which is connected to a $ 120{\text{ V}} $ supply.

Note :
It is good to note that while solving any of the expressions in these types of questions if we have rounded a value up to some decimal places, then the final answer also will get rounded off up to some figure. Also Power, P can be related to current I by $ P = {I^2}R $ , where R is the resistance.