Certainly! Here is the solution rewritten in a more organized manner: --- ## Solution ### First, we determine the states of the diodes: --- #### For D2: - **Anode** at node 1, **cathode** at ground (0V) - Since node 1 is connected via \(R_1\) to \(V_1 = 3V\), node 1 would be \(> 0V\) - Therefore, D2 is **reverse-biased** (OFF) --- #### For D1: - **Anode** at node 1, **cathode** at node 2 - Assume D1 is ON (ideal diode) - Thus, \( V_1 - I R_1 = V_2 \) --- ### Thevenin Equivalent to the Right of D1 - **Looking into the right from node 2:** - The voltage divider consists of \( R_2 = 6k\Omega \) and \( R_3 = 4k\Omega \) - Current across \( R_2 \) (i.e., at node 2) from 3V: \[ V_{Th} = V_2 = \frac{R_2}{R_2 + R_3} V_{in} = \frac{6}{6+4} \times 3 = \frac{18}{10} = 1.8V \] - Now, \[ R_{Th} = R_2 \parallel R_3 = \left(\frac{1}{6000} + \frac{1}{4000}\right)^{-1} = \left(\frac{2+3}{12000}\right)^{-1} = \left(\frac{5}{12000}\right)^{-1} \] \[ = \left(\frac{1}{2400}\right)^{-1} = 2400\Omega = 2.4k\Omega \] --- #### Note: Replace the right side of D1 with its **Thevenin equivalent**. --- ### Then, D1 connects \( R_1 = 12k\Omega \), \( V_1 = 3V \), to the Thevenin circuit with: - \( V_{Th} = 1.8V \) - \( R_{Th} = 2.4k\Omega \) --- ### Now, calculate \( V(2) \) using voltage divider: \[ V(2) = \frac{R_{Th}V_1 + R_1V_{Th}}{R_1 + R_{Th}} \] \[ = \frac{2.4(3) + 12(1.8)}{12 + 2.4} = \frac{7.2 + 21.6}{14.4} \] \[ = \frac{28.8}{14.4} = 2.0V \] --- **Hence,** \[ \boxed{V(2) = 2.0V} \] (Answer for part (a)) ---