\$ V_o \$ in this simple circuit with 2 diodes and 3 batteries

Question

Take \$V_{d (on)}\$ to be 0.6V for simplicity.

What's \$V_o\$? Is it voltage across 1K Resistor?

So... what's the direction of current flow in the 1k resistor? Is it away from BAT3 towards \$V_o\$?

How do two diodes and two batteries (BAT1 and BAT2) being in parallel would affect \$V_o? \$

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I think \$V_o\$ is voltage across R1 which is \$10V - i \cdot 1K - V_d - 9V = 0\$

\$ i_{R1} \cdot 1K = V_{R1} = 10V - V_d - 9V = 10 - 0.6 - 9 = 0.3V \$

KVL loop.

But if BAT3 was 5V, then what would be \$ V_o \$?

I know that in this case, current has to go from diodes towards BAT3, right?

Do I just look at one "branch" to calculate \$ V_o \$? For instance, the branch with upper diode, D1.

KVL loop 1:

So Vo = 9V + 0.6V - 5V ??

Answer 1

Vo is the voltage on that node specifically. The voltage across the resistor will be 10-9-0.6 = 0.4V The voltage Vo, which is the voltage in that node is going to be 9 + 0.6V = 9.6V. As long as BAT3 has a greater voltage than BAT1 or BAT2 , the current will be being sourced, ie, leaving BAT3 towards the circuit. If BAT1 is different of BAT2, the voltage Vo will be with the voltage of the less value betwenn BAT1 and BAT2 +0.6V. So if BAT2 = 6V and BAT1 = 9V, Vo will be 6+ 0.6V = 6.6V

If BAT3 = 5V the voltage in Vo will be 5V too, as there will be no voltage across R3 due to the fact the diodes will be reverse-biased, thus allowing no current to flow through them.