1. **Problem Statement:** Find the output voltage $V_0$ in a circuit with two ideal diodes connected to voltage sources $V_1$ and $V_2$ through 1 ohm resistors, converging at $V_0$ which is connected to ground through a 9 ohm resistor. 2. **Key Concept:** Ideal diodes conduct (act as short circuits) when forward biased and block current (act as open circuits) when reverse biased. 3. **Step 1: Analyze diode conduction for $V_1=10V$, $V_2=5V$** - Diode $D_1$ is forward biased if $V_1 > V_0$. - Diode $D_2$ is forward biased if $V_2 > V_0$. 4. **Step 2: Assume both diodes conduct (both forward biased).** - Then $V_0$ is connected to both $V_1$ and $V_2$ through 1 ohm resistors. - The equivalent circuit is two voltage sources $V_1=10V$ and $V_2=5V$ connected to $V_0$ through 1 ohm resistors, and $V_0$ connected to ground through 9 ohms. 5. **Step 3: Calculate $V_0$ using node voltage method:** Let currents from $V_1$ and $V_2$ to $V_0$ be $I_1$ and $I_2$ respectively. $$I_1 = \frac{V_1 - V_0}{1}, \quad I_2 = \frac{V_2 - V_0}{1}, \quad I_3 = \frac{V_0 - 0}{9}$$ At node $V_0$, sum of currents is zero: $$I_1 + I_2 = I_3$$ Substitute: $$\frac{10 - V_0}{1} + \frac{5 - V_0}{1} = \frac{V_0}{9}$$ Simplify: $$15 - 2V_0 = \frac{V_0}{9}$$ Multiply both sides by 9: $$135 - 18V_0 = V_0$$ $$135 = 19V_0$$ $$V_0 = \frac{135}{19} \approx 7.11V$$ 6. **Step 4: Check diode conduction assumptions:** - For $D_1$: $V_1=10V > V_0=7.11V$ forward biased, correct. - For $D_2$: $V_2=5V < V_0=7.11V$ reverse biased, assumption incorrect. 7. **Step 5: Recalculate assuming only $D_1$ conducts:** - $D_2$ is off, so no current from $V_2$. - Currents: $$I_1 = \frac{10 - V_0}{1}, \quad I_3 = \frac{V_0}{9}$$ At node $V_0$: $$I_1 = I_3$$ $$\frac{10 - V_0}{1} = \frac{V_0}{9}$$ Multiply both sides by 9: $$90 - 9V_0 = V_0$$ $$90 = 10V_0$$ $$V_0 = 9V$$ 8. **Step 6: Check diode conduction:** - $D_1$: $10V > 9V$ forward biased, correct. - $D_2$: $5V < 9V$ reverse biased, correct. So $V_0 = 9V$ for case 1. 9. **Step 7: For $V_1 = -5V$, $V_2 = 5V$** - Check diode conduction: - $D_1$ conducts if $-5V > V_0$ unlikely. - $D_2$ conducts if $5V > V_0$ possible. Assume only $D_2$ conducts. 10. **Step 8: Calculate $V_0$ with only $D_2$ conducting:** $$I_2 = \frac{5 - V_0}{1}, \quad I_3 = \frac{V_0}{9}$$ At node $V_0$: $$I_2 = I_3$$ $$\frac{5 - V_0}{1} = \frac{V_0}{9}$$ Multiply both sides by 9: $$45 - 9V_0 = V_0$$ $$45 = 10V_0$$ $$V_0 = 4.5V$$ 11. **Step 9: Check diode conduction:** - $D_1$: $-5V > 4.5V$ false, off. - $D_2$: $5V > 4.5V$ true, on. So $V_0 = 4.5V$ for case 2. **Final answers:** - For $V_1=10V$, $V_2=5V$, $V_0 = 9V$ - For $V_1=-5V$, $V_2=5V$, $V_0 = 4.5V$