Question

ee 201: electric circuits
from the original circuit: $v_{rl}=$ 0.72 v $p_{rl}$
from the thevenin circuit: $v_{rl}=$ 0.72 v $p_{rl}$
$v_{th}$ calculated 4.59 v
$r_{th}$ 1

Explanation:

Step1: Identify the problem

We need to find the Thevenin resistance ($R_{TH}$) for the given circuit. Since the voltage across $R_L$ and calculations for $V_{TH}$ are already shown, and we need to find $R_{TH}$.

Step2: Remove the load and voltage sources

Remove $R_L$ and replace voltage sources $V_{S1}$ and $V_{S2}$ with short - circuits.

Step3: Calculate equivalent resistances

First, consider the resistors in series and parallel combinations. For the left - hand side part of the circuit with $R_1 = 6.8k\Omega$, $R_2=4.7k\Omega$, $R_3 = 6.8k\Omega$, the equivalent resistance of the parallel combination of $R_2$ and the series combination of $R_1$ and $R_3$ needs to be calculated. Similarly, for the right - hand side part with $R_5 = 2.2k\Omega$, $R_6=1.5k\Omega$, $R_7 = 2.2k\Omega$. After calculating all the parallel and series combinations of resistors in the circuit with voltage sources short - circuited, we get the Thevenin resistance.
Let's calculate step - by - step.
The series combination of $R_1$ and $R_3$ is $R_{13}=R_1 + R_3=6.8k\Omega+6.8k\Omega = 13.6k\Omega$.
The parallel combination of $R_{13}$ and $R_2$ is $R_{123}=\frac{R_{13}\times R_2}{R_{13}+R_2}=\frac{13.6k\Omega\times4.7k\Omega}{13.6k\Omega + 4.7k\Omega}=\frac{63.92}{18.3}k\Omega\approx3.49k\Omega$.
For the right - hand side, the series combination of $R_5$ and $R_7$ is $R_{57}=R_5 + R_7=2.2k\Omega+2.2k\Omega = 4.4k\Omega$.
The parallel combination of $R_{57}$ and $R_6$ is $R_{567}=\frac{R_{57}\times R_6}{R_{57}+R_6}=\frac{4.4k\Omega\times1.5k\Omega}{4.4k\Omega + 1.5k\Omega}=\frac{6.6}{5.9}k\Omega\approx1.12k\Omega$.
The Thevenin resistance $R_{TH}$ is the series combination of $R_{123}$ and $R_{567}$, so $R_{TH}=3.49k\Omega+1.12k\Omega = 4.61k\Omega$.

Answer:

$4.61k\Omega$