I think the easiest method to solve such problems is to assume that the diodes are off (both, and then one of the two), compute the voltages across the diodes and see if there's a contradiction with your assumption. Let's call the top left diode \$D_1\$ and the diode in the middle \$D_2\$.
Case 1: \$D_1\$ off, \$D_2\$ off: Since \$D_1\$ is off there is no current through the top 5k resistor, and since \$D_2\$ is off, there is also no current through the bottom left 10k resistor. So \$V=0\$ and the voltage at the anode of \$D_1\$ is 15 Volts. Contradiction! (\$D_1\$ should be on).
Case 2: \$D_1\$ off, \$D_2\$ on: again no current through top 5k resistor. Voltage \$V\$ is
$$V=\frac{15V\cdot(5k||10k)}{10k+(5k||10k)}=3.75V$$
Contradiction! (Because the voltage across \$D_1\$ would be \$15V-3.75V=11.25V\$ and it should be on.)
Case 3: \$D_1\$ on, \$D_2\$ off: Voltage \$V\$ is
$$V=\frac{15V\cdot 10k}{5k+10k}=10V$$
The voltage at the anode of \$D_2\$ is \$15V\cdot 5k/15k=5V\$. This agrees with our assumption, because with these voltages \$D_2\$ must be off. So your solution is
$$I=0A,\quad V=10V$$
Neither answer is correct, but you mostly had the right idea.
You correctly calculated the current through the whole circuit: \$\dfrac{10 - (-2)}{2000 + 4700} = 1.791 mA\$
The next step you could take would be to calculate the voltage drop across each resistor:
- Across the 2K resistor: 2000 * 0.001791 = 3.58 V
- Across the 4.7K resistor: 4700 * 0.001791 = 8.42 V
To find the voltage at Vo, start at either end and add or subtract the voltage drop across the resistor.
- If we start at the 10V end: Vo = 10 - 3.58 = 6.42 V
- If we start at the -2V end: Vo = -2 + 8.42 = 6.42 V
Best Answer
As shown in the original post, the lower diodes can never be turned OFF because they'll never be reverse biased.
"Ideal" diodes can be likened to switches with zero resistance between the contacts when ON, and infinite resistance between the contacts when OFF.
Redrawing the circuit from that point of view, with perfect switches and with the original post's context allowing us to turn OFF the two bottom diodes, we have: