Question

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

Answer 1

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.