Pulling down voltage to zero on parallel circuit elements

[wesley1]

Hello!

I tried searching for a solution to my problem, but could not find any related topics so far. Perhaps I'm missing the right terms..

There is a simplified circuit like in the attached picture.
I am looking for a method to drive the voltage to zero on R1 with access only to the point in the circuit marked with the question mark.
It is clear that I can control the voltage and current for the rest of the circuit (i.e. R2 and R3).
But what are my options if I want to influence the R1 part of the circuit by putting components where the question mark is..?



I suppose shorting the circuit there is not a great idea. Putting a low value resistor with high wattage to let most current bypass the rest of the components..?

Thanks!

[Zero999]

The only way to force the voltage across R1 to 0V is by short circuiting it, which will cause a large current to flow.

What does the circuit do? Can we please see the bigger picture.

[macboy]

Nothing you can place at the question mark can affect the voltage across R1 in any way... not without knowing more about the 12 V source at least. If it is an "ideal" 12 V source, with zero internal resistance, and the wires are also ideal, then no amount of load will drag it down. Since the 12 V source is connected directly across R1, the voltage at R1 is simply exactly equal to that source voltage, period. Nothing you can place at the question mark can create a short across R1 or across the 12 V source. Even placing a voltage source at the question mark can't change the voltage across R1.
You said it's a simplified circuit. Has it been "simplified" correctly? Is everything you know about the circuit (especially the 12 V source) shown? Unless something is wrong or missing, there is no solution.

[wesley1]

It is an intercom circuit with a proprietary one wire protocol. The master provides the 12V to the slave devices in the appartments which use parasitic power from the line.
There is a two-way communication between the master and the slaves.

I'd like to experiment with arduino to be able to trigger my intercom without the real master.
I assumed the slaves pull the signal level to ground for sending. Maybe my assumption was wrong and they reply by some different technique.

One thing to note is that I have access also to GND at the question mark, R2 is the device in my appartment.
About the voltage source, it's a 12VDC power supply driven by a 230VAC to 12VAC transformer with a 100mA fuse. That's everything I could figure out from the docs.

[Kirill V.]

Firstly, you should consider internal resistance of 12 V source and add it to your simplified circuit

[rstofer]

To get the voltage to 0V requires 0 Ohms given that there is a source voltage.  V = I * R and if we divide through by R, we get V/R = I and, since division by 0 is undefined, we can assume the current approaches infinity.  So, for any non-zero voltage, with 0 Ohms, we get infinite current.

Ohm's Law still applies, it's the division by 0 that causes the problems.  One solution is to model an internal resistance for the battery.  The other is to realize it is a trick question since an internal resistance wasn't given and, from Ohm's Law, infinite current must be supplied.

I'm guessing that they want the simple answer:  To get 0V across the resistor means you must short circuit the resistor OR remove it from the circuit. 

[tpowell1830]

hi Wesley1, your assumption that the supply is a 12 VAC supply that supplies 12 VDC is a bit suspicious. A transformer indeed provides AC voltage/current, however, in order to calculate the rectified DC voltage, assuming that it gets rectified in the normal way with diodes, will yield an average DC voltage that is somewhat higher than 12 volts. The resistive loads that you you are showing are all in parallel, which yields a current that is the parallel resistance of all 3 resistors. From your description, there is not enough information to evaluate this circuit and make any conclusions. One thing is for sure, you cannot pull the voltage to zero at the point of the question mark, without either blowing the fuse or simply overloading the supply. The series resistive equivalent, that has been pointed out by several others, will determine whether the current gets high enough to blow the 100 mA fuse. I doubt that the equivalent series resistance is that high, but it is possible.

You also stated that the power supply was fused to 100 mA, which means that the maximum current can only be 100 mA. If you short this system anywhere, the current will quickly blow the small fuse.

This leads me to question whether you have drawn this circuit correctly, as most pulldown resistor configurations would be in series with ground. Perhaps you have misunderstood/misrepresented the circuit?

!!!!**ADVICE WARNING DETECTOR**!!!!
On another note, it is always a good idea to state your problem in terms of your goals, rather than asking an open ended question as you did, however, you did that in a later post, somewhat. I also suggest that you study Ohm's law and Thevenin and Norton's theorems, which will quickly show you that your original premise is unlikely.
****!!ADVICE SECTION COMPLETED!!****

[Nerull]

Your suggestion is to short out the power supply with a battery, potentially damaging two components instead of one?

That has zero advantages over a direct short. You've just made a battery or powersupply into an expensive piece of wire, since you need a low internal resistance for it to pull down the voltage close to 0. Except its even more likely to burst into flames.

EDIT: This post makes no sense now because the post it was in response to was deleted.

[wesley1]

Thanks for the replies!

I think it's better to redraw the circuit like this:
Measuring voltage across question mark shows 13.26 V



Theoretically if the voltage source has some decent overcurrent protection, I could short the circuit with an optocoupler for 300 microsecs to 1.3 millisecs the longest..?
Like with this circuit:


Attached some pics from the communication on the 12V line with scope and logic analyzer: