Electrical – the function of the zener diodes used in this setup

Consider the operation of the circuit.

When the transistor is on current is flowing in the coil from top to bottom as the circuit is drawn we now switch the transistor off. The current in the coil still wants to flow.

For the circuit on the left this current can now flow back to Vcc via the diode the voltage across the coil has reversed direction and is limited by the diode the current can decay to zero safely.

For the circuit on the right the diode does not help. The current flowing in the coil will force the voltage on the collector to rise to the point where the transistor (or possibly the diode) breaks down and starts to conduct. At this point the current can start to decay in the coil but the energy in the broke down transistor (or less likely diode) will be excessive and may well result in the transistors death. Note a zener diode here will work because you allow the voltage on the coil to reverse so the current can decay to zero while limiting the voltage across the transistor to a safe value.

It should be noted the allowing the voltage across the coil to reverse to an higher voltage means the current can decay more quickly which is why you sometimes see a zener in the right hand circuit or more than one diode in series in the left hand one.

I'd start with the assumption that both diodes are forward biased and see what happens.

D1 connected to ground means the anode is 0.7V. D2 will have a forward drop of 0.7 meaning that it's actually creating a virtual ground at V(out).

Then it's just voltages over resistors. (5-0.7)/10k = current through 10k. (0-(-5))/5k = current through 5k.

Then use KCL to determine what I is. Current through 10k flows into node, current through 5K flows out of node, and current I flows out of node. So you should have an equation like this:
$$I_{10k}-I_{5k}-I=0$$

When you do this, you'll see that I is actually negative which can't be because the diode would then be reversed biased and would block all current. That means D1 becomes an open in this circuit.

Now you just have one series line of voltage across resistors and a diode. Ohms law states this:
$$\frac{5-(-5)-0.7}{10k+5k}=I_{series}$$

With Iseries and 5k, you can find Vout. You know that I is 0 because D1 is reverse biased.