find the current flowing out of the battery. 23.9 ω 17.0 ω 43.7 ω 12.0 v i = ? a

Explanation:

Step1: Calculate parallel - resistance of 17.0Ω and 43.7Ω

The formula for parallel - resistance $R_{p}$ of two resistors $R_1$ and $R_2$ is $R_{p}=\frac{R_1\times R_2}{R_1 + R_2}$. Here, $R_1 = 17.0\Omega$ and $R_2=43.7\Omega$. So, $R_{p}=\frac{17.0\times43.7}{17.0 + 43.7}=\frac{742.9}{60.7}\approx12.24\Omega$.

Step2: Calculate total resistance of the circuit

The total resistance $R_{total}$ of the circuit is the sum of the series - resistance and the parallel - resistance. The series resistance is $R_s = 23.9\Omega$ and the parallel resistance is $R_{p}\approx12.24\Omega$. So, $R_{total}=R_s+R_{p}=23.9 + 12.24=36.14\Omega$.

Step3: Use Ohm's law to find the current

Ohm's law is $I=\frac{V}{R}$, where $V$ is the voltage and $R$ is the resistance. Given $V = 12.0V$ and $R = R_{total}=36.14\Omega$. So, $I=\frac{12.0}{36.14}\approx0.33A$.

Answer:

$0.33$