Question

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

• A. $22.5A$
• B. $2.25A$
• C. $0.67A$
• D. $6.7A$

Answer 1

The correct option is C $0.67A$

The answer is C.
Components connected in series are connected along a single path, so the same current flows through all of the components. The current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component. In a series circuit, every device must function for the circuit to be complete. One bulb burning out in a series circuit breaks the circuit. A circuit composed solely of components connected in series is known as a series circuit. The total resistance of resistors in series is equal to the sum of their individual resistances. That is, $R_{total}=R_1+R_2...R_n$.
In this case, the total resistance is $0.5+0.3+0.4+0.2+12=13.4ohms$.
The total current in the circuit is calculated from the relation $V=IR$, that is, $I=V/R.$
$I=\frac{9V}{13.4ohms}=0.67A.$
As it is a series connection, the current through each of the components is the same.
Hence, the current through the $0.4$ resistors is $0.67A$.