What current flows through the bulb of a \({\bf{3}}{\bf{.00}}\;{\bf{V}}\) flashlight when its hot resistance is \({\bf{3}}{\bf{.60}}\;{\bf{\Omega }}\) ?

The given data can be listed below as,

• The potential difference across the flashlight is,\(V = 3.00\;{\rm{V}}\).
• The hot resistance of the flashlight is, \(R = 3.60\;{\rm{\Omega }}\).

In this question, the electric current through the bulb directly depends on the factors like the resistance of the flashlight and potential differences across the light.

According to Ohm’s law, the expression to calculate the current flows through the bulb is expressed as,

\(\begin{align}{}V &= IR\\I & = \frac{V}{R}\end{align}\)

Here,\(I\)is the current flow through the bulb.

Substitute all the known values in the above equation.

\(\begin{align}I & = \left( {\frac{{3.00\;{\rm{V}}}}{{3.60\;{\rm{\Omega }}}}} \right)\\ & \approx 0.833\;{\rm{V/\Omega }}\\ & \approx \left( {0.833\;{\rm{V/\Omega }}} \right)\left( {\frac{{1\;{\rm{A}}}}{{1\;{\rm{V/\Omega }}}}} \right)\\ & \approx 0.833\;{\rm{A}}\end{align}\)

Thus, the current flow through the bulb is \(0.833\;{\rm{A}}\).