Question

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

Answer 1

There is no current division occurring in a series circuit. Current flow through the component is the same, given by Ohm’s law as
$V= IR$
$I=\frac VR$
Where,
R is the equivalent resistance of resistances 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. These are connected in series. Hence, the sum of the resistances will give the value of R.
R = 0.2 + 0.3 + 0.4 + 0.5 + 12
R = 13.4 Ω
Potential difference, V= 9 V
I = $\frac {9}{13.4}$
I = 0.671 A

Therefore, the current that would flow through the 12 Ω resistor is 0.671 A.