3 Tutor System
Starting just at 265/hour

# 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.