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: 
(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).
This circuit converts a current at its input to a voltage at its output. It is typically used for digital signals, and with a base resistor at the first transistor. (In that case, if the base resistor is larger than RL, the output impedance is lower than the input impedance, i.e., the circuit amplifies the current.)

Ue and Ua,2 are not actually interesting, because both transistors are either off or saturated.

The capacitor, if it has been put there deliberately and does not just represent a parasitic capacitance, slows down switching on the second transistor. This might be useful if you want to balance the switching times (a saturated transistor is slower switching off than switching on).
For example, such a capacitor is used in several Roland MIDI devices to ensure that the falling and raising edges of the MIDI signal are delayed by approximately the same amount:  

![Roland MKS-30 MIDI I/O](https://i.stack.imgur.com/B7f1K.png)