Answer
0.35 A
0
Work Step by Step
The 3 resistors are in series. When resistors are wired in series, we use equation 19–3 to find the equivalent resistance.
$$R_{eq}=2.0\Omega+9.5\Omega+14.0\Omega=25.5\Omega$$
The current in the entire circuit is the battery voltage divided by the equivalent resistance: $I=\frac{\epsilon}{R_{eq}}=\frac{9.0V}{25.5\Omega}=0.3529A\approx0.35A$.
Now sum up the voltage changes around the circuit. Start at the negative terminal of the battery and proceed counterclockwise.
$$\Sigma \Delta V=9.0V-(0.3529A)(9.5\Omega) -(0.3529A)(14.0\Omega) -(0.3529A)(2.0\Omega)=0$$
