Question: I have a large power supply with multiple DC output modules. 2 of the modules are connected, positive to positive, through a 1.1 ohm resistor (really, 2 2.2 ohm resistors in parallel). When I measure the voltages without the resistor, I have the correct voltages (3.65V and 5.35V) but with the resistor I measure 5.9V on the 5.35V module, and 2.8V on the 3.65V module. I am probing across the + and common terminals of each module. I was also curious as to see if the commons were corrected, and when measured, it reads 67.5 ohms (However, inside the machine when all hooked uo, the commons become connected.) So I suppose my circuit would look a little something like this (outside of the machine)...

![schematic](https://i.stack.imgur.com/SgLPb.png)

[simulate this circuit](/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fSgLPb.png) – Schematic created using [CircuitLab](https://www.circuitlab.com/)

My question is, what is the exact purpose of this circuit (My guess is to share the current for the loads) Also note, the resistors are only connected to the 2 positive output terminals. And how should I properly measure the voltages?[![enter image description here](https://i.stack.imgur.com/mMCF2.jpg)](https://i.stack.imgur.com/mMCF2.jpg) (This goes to an Ultrasound Medical Device, the computer reads the voltages correctly, but am I probing wrong? Red probe is indicating the 5.35V module and black is 3.65V. The blue line coming off the terminals is the power resistors located somewhere else.)

Answer: 
Given your instructor's remark, I suspect that you were actually told about "debouncing", which a very useful function. However, the circuit shown is a bad, bad example of how to do it.

Let's think about a switch making contact. Inside, you have two pieces of metal making contact, and on a timescale of milliseconds the two pieces can actually bounce, perhaps repeatedly, before settling down. If this happens the switch output will apparently show multiple activations where only one was intended. To combat this, a circuit like this can be used:

![schematic](https://i.stack.imgur.com/bzSBf.png)

[simulate this circuit](/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fbzSBf.png) – Schematic created using [CircuitLab](https://www.circuitlab.com/)

You've noticed that it has the same general outline as your circuit, but a few more resistors. 

Let's say that the switch is open, and gets briefly closed, then open, then closed for good (a single bounce). Current through R1 turns on Q1 and pulls the collector low. This also discharges C1. As a result Q2 is turned off and Vout goes high. When the switch bounces, Q1 is briefly turned off and its collector rises. However, the resistor charges up much more slowly than it discharges (for appropriate resistor values of R2 and R3), so Q2 never turns back on. As a result, the output Vout has been "debounced", and R2/R3/C1 can be selected for any desired bounce time to be ignored.

The original circuit is not very good, since the capacitor voltage swing is quite small, due to the clamping effect of the Q2 base-emitter junction.
(comments to answer)

*The workings of the "circuit"*

The voltage at the collector of the first BJT will be clamped at 0.7V (approx) by the base-emitter junction of the second BJT. BJT(2) is most likely in saturation (so only a small DC output say 0.2V). BJT(1) has no bias arrangement so it impossible to tell what Ue will actually do to BJT(1). (Is Ue a DC or AC or mixed signal? - no one knows. ). **Its a terrible example to give to a class who are just learning to come to grips with the basics.** I'm not surprised you were confused.

*What does the capacitor achieve in this circuit ?*

Very little and I can't understand why anyone would put it across BJT(1). As the voltage across it is clamped (by BJT(2)) all it can do is act as a 'smoothing' capacitor rather than a coupling capacitor (collector to base). If Ue is taken to +0.7V then BJT(1) would be in saturation (Vc-e) = 0.2V causing the voltage across the capacitor to fall (discharging rapidly through BJT(1)). In this case Ua2 goes high because BJT(2) is turned off. If Ue is taken low (0V) the voltage across the capacitor rises to 0.7V with the time constant = RL\*Cp and the Ua2 falls to Vsat (about 0.2V).