# A battery of 9 V is connected with resistors

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

solution:
Since all the resistors are in series, equivalent resistance,

$R_{\varepsilon}=0.2+0.3+0.4+0.5+12=13.4 \Omega$

Current, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{2}}=\frac{9}{13.4}=0.67 \mathrm{~A}$

In series, same current (I) flows through all the resistors. Thus, current flowing through $12 \Omega$ resistor $=0.67 \mathrm{~A}$