Question

A battery with a terminal voltage of 12V is connected to a circuit consisting 

of four 10 Ω and one 5 Ω resistor all in series. Assume the battery has 

negligible internal resistance. Calculate the current through each resistor.

Answer 1

Answer

Equivalent resistance is

$R=4*10+5=45\Omega$

These resistances are in series so current will be same.

$I=\frac{V}{R}\\=\frac{12}{45}\\=0.27A$