Question

consider the network of four resistors shown in the diagram, where $r_1 = 2.00 omega$, $r_2 = 5.00 omega$, $r_3 = 1.00 omega$, and $r_4 = 7.00 omega$. the resistors are connected to a constant voltage of magnitude $v$. (figure 1)
part a
find the equivalent resistance $r_{eq}$ of the resistor network. express your answer in ohms.
part b

Explanation:

Step1: Calculate parallel - resistance of $R_1$ and $R_2$

The formula for two - resistors in parallel is $R_{12}=\frac{R_1\times R_2}{R_1 + R_2}$. Substituting $R_1 = 2.00\Omega$ and $R_2=5.00\Omega$, we get $R_{12}=\frac{2\times5}{2 + 5}=\frac{10}{7}\Omega$.

Step2: Calculate total series - resistance

The resistors $R_{12}$, $R_3$, and $R_4$ are in series. The formula for resistors in series is $R_{eq}=R_{12}+R_3 + R_4$. Substituting $R_{12}=\frac{10}{7}\Omega$, $R_3 = 1.00\Omega$, and $R_4=7.00\Omega$, we have $R_{eq}=\frac{10}{7}+1 + 7=\frac{10+7 + 49}{7}=\frac{66}{7}\approx9.43\Omega$.

Answer:

$9.43$