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A current of $0.5\,A$ is drawn by a filament of an electric bulb for $10$ minutes.Find the amount of electric charge that flows through the circuit.

Answer
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Hint:Here we will first know about the concept of electric charge carrier, relation between charge, current and time; and then by using that relation we will solve the given problem.A property of matter is electric charge. Positive and negative electric charges are the two types of electric charge that are commonly used.

Formula used:
The equation relating electric charge, current and time is:
$Q=I\times T$
Where $Q$ is the electric charge, $I$ is the electric current, and $T$ is the time.

Complete step by step answer:
When two objects with the same charge interact, they repel each other. Attractive forces exist between two objects with opposite charges. The symbol for electric charge is $Q$, and it is calculated in coulombs ($C$).Since electric charge is quantised, every object can only have an integer multiple of a given charge value. The charge magnitude of an electron, which is roughly $1.6\times {{10}^{-19}}C$, is the smallest value of magnitude of charge an object can have.

Negatively charged electrons. A neutral material becomes more positively charged as it loses electrons. A neutral material becomes more negatively charged as it gains electrons. The rate of flow of positive charge is known as current. The movement of electrons, ions, or other charged particles may trigger current. Since electrons are negatively charged, they flow in the opposite direction of current. We know that,
$Q=I\times T$
\[electric~charge=electric~current\times time\]

Electric current flows from a power supply's positive terminal to the negative terminal or field, or from the ground to a negative terminal in an electric circuit. Electric current is denoted by the letter $I$ and is measured in amperes ($A$). Here the given parameters are:
Current (I) = 0.5 A
Time = 10 minutes
Time in minutes has to be converted into seconds as follows
\[Time\text{ }\left( t \right)\text{ }=\text{ }10~\times \text{ }60\text{ }=\text{ }600\text{ }sec\]
$Q=I\times T$
$\Rightarrow Q=0.5\times 600$
$\therefore Q=300\,C$

Therefore, 300 C is the amount of charge flowing through the circuit.

Note:One must remember to change the units into SI units, like here time was given to be 10 minutes but we had to convert it to seconds in order to get the correct answer, if conversion of the unit is forgotten one might end up having an incorrect answer.