A total of \({\bf{600}}\;{\bf{C}}\) of charge passes through a flashlight in\({\bf{0}}{\bf{.500}}\;{\bf{h}}\). What is the average current?

The given data can be listed below as,

• The total electric charge is,\(Q = 600\;{\rm{C}}\).
• The time interval is, \(t = 0.500\;{\rm{h}}\).

The value of the average current can be calculated from the value of the charge quantity that flows in the given time. Generally, the measuring unit of the current is Ampere (A).

The relation to calculate the average current is expressed as,

\(\begin{aligned}Q = It\\I = \frac{Q}{t}\end{aligned}\)

Here,\(I\)is the average current.

Substitute all the known values in the above equation.

\(\begin{aligned}I = \left( {\frac{{600\;{\rm{C}}}}{{0.500\;{\rm{h}}}}} \right)\left( {\frac{{\frac{1}{{3600}}\;{\rm{C/s}}}}{{1\;{\rm{C/h}}}}} \right)\left( {\frac{{1\;{\rm{A}}}}{{1\;{\rm{C/s}}}}} \right)\\ \approx 0.333\;{\rm{A}}\end{aligned}\)

Thus, the average current is\(0.333\;{\rm{A}}\).