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 \(\Omega \) resistor?

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.