1. **Problem Statement:** Determine the Thevenin equivalent voltage ($V_{TH}$) and Thevenin equivalent resistance ($R_{TH}$) seen from terminals A and B when a 68 Ω resistor is connected in parallel across $R_2$ and $R_3$. 2. **Given Data:** - $V_S = 12V$ - $R_1 = 100\ \Omega$ - $R_4 = 120\ \Omega$ - $R_2 = 68\ \Omega$ - $R_3 = 20\ \Omega$ - Additional resistor $R_p = 68\ \Omega$ connected in parallel with $R_2$ and $R_3$ 3. **Step 1: Find the equivalent resistance of $R_2$, $R_3$, and $R_p$ in parallel.** The combined parallel resistance $R_{eq}$ is given by: $$ \frac{1}{R_{eq}} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_p} $$ Substitute values: $$ \frac{1}{R_{eq}} = \frac{1}{68} + \frac{1}{20} + \frac{1}{68} = \frac{1}{68} + \frac{1}{68} + \frac{1}{20} = \frac{2}{68} + \frac{1}{20} $$ Calculate: $$ \frac{2}{68} = \frac{1}{34} \approx 0.02941, \quad \frac{1}{20} = 0.05 $$ So: $$ \frac{1}{R_{eq}} = 0.02941 + 0.05 = 0.07941 $$ Therefore: $$ R_{eq} = \frac{1}{0.07941} \approx 12.59\ \Omega $$ 4. **Step 2: Find total resistance seen by the source.** The circuit from the source is $R_1$ in series with two branches: one branch is $R_4$ and the other is $R_{eq}$. These two branches are in parallel, so their combined resistance $R_p'$ is: $$ \frac{1}{R_p'} = \frac{1}{R_4} + \frac{1}{R_{eq}} = \frac{1}{120} + \frac{1}{12.59} $$ Calculate: $$ \frac{1}{120} = 0.00833, \quad \frac{1}{12.59} = 0.07941 $$ Sum: $$ 0.00833 + 0.07941 = 0.08774 $$ So: $$ R_p' = \frac{1}{0.08774} \approx 11.40\ \Omega $$ Total resistance $R_{total}$ is: $$ R_{total} = R_1 + R_p' = 100 + 11.40 = 111.40\ \Omega $$ 5. **Step 3: Calculate total current from the source.** Using Ohm's law: $$ I = \frac{V_S}{R_{total}} = \frac{12}{111.40} \approx 0.1077\ A $$ 6. **Step 4: Calculate voltage at node A (across $R_4$ and $R_{eq}$ parallel combination).** Voltage drop across $R_1$: $$ V_{R1} = I \times R_1 = 0.1077 \times 100 = 10.77\ V $$ Voltage at node A (also $V_{TH}$) is: $$ V_{TH} = V_S - V_{R1} = 12 - 10.77 = 1.23\ V $$ 7. **Step 5: Calculate Thevenin resistance $R_{TH}$.** Deactivate the voltage source (replace $V_S$ with a short circuit). From terminals A and B, $R_1$ is in series with the parallel combination of $R_4$ and the parallel of $R_2$, $R_3$, and $R_p$. We already calculated the parallel of $R_2$, $R_3$, and $R_p$ as $12.59\ \Omega$ and the parallel of $R_4$ and $R_{eq}$ as $11.40\ \Omega$. So: $$ R_{TH} = R_1 + R_p' = 100 + 11.40 = 111.40\ \Omega $$ **Final answers:** $$ V_{TH} \approx 1.23\ V $$ $$ R_{TH} \approx 111.4\ \Omega $$ These values do not match the options exactly, so rechecking the problem setup or options might be necessary.