Question

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

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
$I=\frac{V}{R}$
where, R is the equivalent resistance of resistances $0.2\Omega ,0.3\Omega ,0.4\Omega ,0.5\Omega$ and $12\Omega$. These are connected in series.
Hence, the sum of the resistances will give the value of R.
$R=0.2+0.3+0.4+0.5+12=13.4\Omega$
Potential difference, V = 9 V
I = $\frac{9}{13.4}$ = 0.671 A
Therefore, the current that would flow through the $12\Omega$ resistor is 0.671 A.