To solve this problem, we need to analyze the circuit's behavior with an ideal diode for values of \( V1 \). Circuit Analysis 1. **Idealode Characteristics**: - Conduct (acts like short circuit) forward-biased (positive voltage across). - Does not conduct ( like an open circuit) when reverseased (negative across it). . **Given**: - R1 = 1 \,text{k}\Omega \) ### Completion For each value of \( V1 \): - **\( V1 = -10 \, \text{V} \)**: - The diode is reverse-biased. - The current \( I = \, \text{mA} \). - **\( V1 = -5 \, \text{V} \)**: - The diode is reverse-biased. - The current \( I = \, \text{mA} \). - **\( V1 = \, \text{V} \)**: - The diode is neither forward- nor reverse-biased, but no potential for conduction. - The current \( I = \, \text{mA} \). - **\( V1 = +5 \, \text{V} \)**: - The diode is forward-biased. - The current \( I = \frac{V1}{R1} = \frac{5}{100} = 5 \, \text{mA} \). - **\( V1 = +10 \, \text{V} \)**: - The diode is forward-biased. - The current \( I = \frac{V1}{R1} = \frac{10}{100} = 10 \, \text{mA} \). ### Completed Table | \( V1 \) (V) | \( I \) (mA) | |--------------|--------------| | -10 | | | -5 | | | | | | +5 | 5 | | +10 | 10 |