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 b two resistors of resistance $r_5 = 3.00 omega$ and $r_6 = 3.00 omega$ are added to the network, and an additional resistor of resistance $r_7 = 3.00 omega$ is connected by a switch, as shown in the diagram. (figure 2) find the equivalent resistance $r_b$ of the new resistor network when the switch is open. express your answer in ohms. part c find the equivalent resistance $r_c$ of the resistor network described in part b when the switch is closed.

Explanation:

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

For two resistors in parallel, the formula is $R_{12}=\frac{R_1\times R_2}{R_1 + R_2}$. Given $R_1 = 2.00\Omega$ and $R_2=5.00\Omega$, we have $R_{12}=\frac{2\times5}{2 + 5}=\frac{10}{7}\Omega$.

Step2: Calculate total series - resistance

The resistors $R_{12}$, $R_3$, $R_4$, $R_5$ and $R_6$ are in series when the switch is open. $R_3 = 1.00\Omega$, $R_4=7.00\Omega$, $R_5 = 3.00\Omega$, $R_6=3.00\Omega$. The equivalent resistance $R_B$ is $R_B=R_{12}+R_3 + R_4+R_5+R_6$. Substitute the values: $R_B=\frac{10}{7}+1 + 7+3+3=\frac{10}{7}+14=\frac{10 + 98}{7}=\frac{108}{7}\approx15.43\Omega$.

Answer:

$\frac{108}{7}\Omega$