Question

the circuit below has three resistors. if the current exiting the battery is 0.50 a, what is the potential difference across r3? r1 = 2.0ω, r2 = 3.0ω, r3 = 2.0ω. 1.0 v 0.60 v 0.30 v 1.6 v

Explanation:

Step1: Analyze current division in parallel - part

First, find the equivalent resistance of $R_1$ and $R_2$ in parallel. The formula for two - resistors in parallel is $R_{12}=\frac{R_1\times R_2}{R_1 + R_2}$. Here, $R_1 = 2.0\Omega$ and $R_2=3.0\Omega$, so $R_{12}=\frac{2\times3}{2 + 3}=\frac{6}{5}=1.2\Omega$.

Step2: Analyze the total circuit

The resistors $R_{12}$ and $R_3$ are in series. The total resistance of the circuit $R_{total}=R_{12}+R_3$. Since $R_{12}=1.2\Omega$ and $R_3 = 2.0\Omega$, $R_{total}=1.2 + 2.0=3.2\Omega$.

Step3: Use Ohm's law for the whole circuit

The current in the circuit $I = 0.50A$. According to Ohm's law $V = IR$. The voltage of the battery $V=I\times R_{total}=0.5\times3.2 = 1.6V$.

Step4: Find voltage across $R_3$

Since $R_{12}$ and $R_3$ are in series, the current through $R_3$ is the same as the total current $I = 0.50A$. Using Ohm's law $V_3=I\times R_3$. Substituting $I = 0.50A$ and $R_3=2.0\Omega$, we get $V_3=0.5\times2=1.0V$.

Answer:

$1.0V$