I'm working through a circuit analysis text on my own for the purposes of getting started with Arduino. I just came across a box that explained why putting two different voltage sources in parallel is a bad idea but the explanation didn't quite make sense and now it's bugging me.  I'm aware there is another, similar question here:  but the answers aren't quite what I'm looking for.  What I want to know is: how do I calculate the voltages, currents and resistances in a circuit that clearly defies idealizations? For instance, how much current will my poor different-voltage batteries encounter, respectively?
67804
I have a computer power supply that I'm hacking together as a bench supply. For this particular model to power on, I need a minimum load across both +5V and +12V.  "Easy," I thought, "I'll just connect both +5 and +12 to my power resistor!" And it worked, but then I started thinking, what does it mean to have these two different voltages connected in parallel? If the voltages were the same, then I would be . But what about different voltages?  Also, what if I connected +5 and +12 in series, and then put a load on that? The equivalent voltage would be +17V; what would be the difference between that and parallel?  Or am I going about this the wrong way; should I put a separate resistor on each rail? It seems like I can do better than that.
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I'm curious, when connecting my 2N2222 transistor's base to a PWM output pin on my Arduino, I know the Arduino can deliver a maximum of 40mA current. The 2N2222 only requires around 5 to 10 mA on it's base to switch the collector and emitter. Now, when calculating the base-resistor under "normal" conditions where I'm merely switching it directly from Vcc, the following applies:   However, when putting in a 500 ohm or 1K resistor in (1K for 5mA) directly it works great. But putting it in between the Arduino and the transistor, the transistor doesn't switch.  I suspect thus, that the resistance is too high of the base-resistor. Therefore, we may very well not need a base-resistor.  But, according to spec, the Arduino can deliver 40mA. So, am I correct in assuming the folling: 1) I should calculate the perceived resistance that the Arduino is giving, if it goes high to 5Volt and only gives 40mA, and then 2) Subtract that from the 500ohm resistor needed to make the transistor operate optimally.  Example:    So, I'm theorising that I have to then put in a 375 ohm resistor in to make the total resistance between the Arduino and the base resistor 500 ohm.  Somehow this doesn't feel right. Keep in mind, I'm not an electronic engineer, so I might have the cat by the tail here :P
59450
Is it the current you have to drive in the input to get a specific, defined voltage at the output? And don't the currents cancel each other when applied to both inputs?
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Basically the circuit below is what my professor gave us and we need to solve for the the voltage across the 8 ohm resistor, the current through the 2 ohm resistor, power absorbed by the 10 ohm resistor, and the power delivered by the 4 a source. I used regular mesh and I got the wrong values(I checked using KCL and KVL). One thing that I think I got wrong is I assumed that I1 = 3A(feedback). I tried to search the web for a solution and I've seen something called "super mesh". Is this helpful? If yes, can anyone assist me on how to work with it?  Other thing is I want to check if my values are correct using MIT's Circuit Sandbox, however I don't know where should I put the ground so that the circuit will simulate in the way I wanted it to be. Every help is very much appreciated Thanks!
101840
Have set up a voltage divider inputting 1 volt into a LM741 op amp to act as a voltage follower. For some reason I get 2 volts at the output. Have checked everything I can think of and can’t figure out why.
30815
If I have a plot of current ($y$ axis) against 1/Resistance ($x$ axis).  The circuit it is measured from is a simply 2 resistors connected in parallel to battery, where the potential across the resistors is $1V$. One resistor remains fixed, and the second varies, the resistance plotted on the graph is only the resistance of the second resistor.   Does this mean that the gradient will be voltage, but will it be $1V$ or less, as the resistance on the graph is only one from one of the resistors?  Also what will the $y$ intercept signify? $I = V/R +c$, what does the $c$ represent?
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What's the purpose of the two capacitors in parallel on each side of the regulator in      I've seen similar setups in other similar circuits and can guess that it's related to one being polarized an one not, but I don't really understand what's going on there.
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If I had a 12V battery and wanted to use an LED that used 2V and 20mA, then I need a resistor that will drop 10V and provide enough resistance to produce a current of 20mA. Therefore I would use R=V/I = R = 10/0.020 = 500 Ohms.  Therefore I would need a 500 Ohm resistor. I believe this is correct so far?  My confusion comes from when we then look at the circuit as a whole. The circuit has a current of 0.020A and a resistance of 500 Ohms, which when looking at V=IR equals 10V (the resistor), not the 12V that it should be there for the whole circuit?  Where am I going wrong? How do I make V = IR work regarding the whole circuit so that V = 12?  I'm pretty new to this sort of thing and no doubt I've probably made a daft mistake but if someone could just help clear up my confusion it would be appreciated.  Thanks.
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According to , if there is a potential difference, $V$, across a resistor then there is a current, $I$, flowing through it.   Since we assume that points along the connecting wire are at the same potential, how can current, $I$ flow between points at the same potential, $V$?
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How can I connecting two different DC power sources together? for example 15Vdc(5Amp) with 5Vdc(2Amp). I want to make one DC source (20Vdc,7Amp) by more powerful supplies.
91933
As fa as I understand it, two identicle batteries in parallel will give double the amperage than just one battery and two batteries in series will give double the voltage. If my understanding is correct, what is the reason for this.  Any help is appreciated  source:
96843
I am an electronics beginner and I created a small little circuit, just a battery a resistor and a led, then to ground. This is the circuit:         &ndash; Schematic created using   Now I have a few questions about it. I know voltage is the different of electrical charge between two points, so I measured that difference:  Point A to B: 3.27V Point B to C: 1.82V   First question: Does that mean that the actual voltage of: Point A: 5V Point B: 1.73V Point C: 0V?   So I am not sure if my calculations of voltage are correct. I also have this other questions:   Does the resistor have to be before the component I am trying to protect, like an LED. I have seen people place it after, but wouldn't that give a full 5V to the led or component? Ground has a neutral charge right? so why do electrons want to travel so badly to a point that has no electric charge?   Thank you
107456
Two or more capacitors are said to be connected in parallel if each one of them is connected across the same two points. In a parallel combination of capacitors potential difference across each capacitor is same but each capacitor will store different charge. Why is this true? Why is the potential difference across each capacitor in parallel the same and why will each capacitor store different charge?  Also, in series, why are Potential difference across each capacitor different, while charge is the same?
65138
If I have a configuration of 5 resistors as follows:        &ndash; Schematic created using   How can I calculate the expected total resistance between A and B, given only values for R1-R5?  edit: okay, here's what I've tried so far.  I intended to find the expected voltage drops across each resistor, and from there I knew I could easily find the total resistance of the circuit.  First, I observed three loops in the diagram: R1,R3,R4; R3,R5,R2; and R1,R2,R5,R4.  I remember from  my old electronics class that the voltage drop around each loop in a circuit should always equal zero, as follows: $$ v1+v2-v4-v5=0\\ v1+v3-v4=0\\ v3+v5-v2=0 $$ (Assuming that current is flowing from A to B, I further assume that it is flowing down through r3.  If this is wrong, then the value for v3 would just have the opposite sign, and all would be well).  The sum of currents flowing into a point should equal the current flowing out, so I have these equations as well: $$ \frac{v1}{r1}=\frac{v2}{r2}+\frac{v3}{r3}\\ \frac{v5}{r5}=\frac{v3}{r3}+\frac{v4}{r4}\\ \frac{v1}{r1}+\frac{v4}{r4}=\frac{v2}{r2}+\frac{v5}{r5} $$  So now I have  6 equations in 5 unknowns... and if these equations are correct,  then there should be one redundant equation.  However, plugging them into a cas such as maple cannot find any unique values for v1-v5 in this system (an infinite number, in fact), suggesting that there are at least 2 redundant equations above.  Does this mean that there is no unique resistance for this circuit fragment, or what have I done wrong?
76847
I am powering an Arduino using a 5v battery (3.7v to a step-up module), and I need to measure the voltage. Let's say the battery delivers high current, is it safe to measure the voltage via Analog In? As in this:    The reason I'm asking is that I don't know much about the Arduino's ADC architecture and limitations. So normally, I'd do so for safety:    Are the connections in the first diagram safe for the ADC? Thanks in advance!  Note: A similar question was asked here: (), but it doesn't answer the question of high current loads on the battery.
55110
In an ideal voltmeter there is infinite resistance so because of which no current is flowing through it and the voltage reading is not affected.  The ohm's law stated that      V = I * R   In an ideal voltmeter I = 0 so V = 0 but the voltmeter does show some reading which defies the ohm's law. Please tell me how is this possible.
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I simply can't figure out what the capacity of this circuit is.  I can do the math myself, I just need a hint how to create an equivalent circuit where it is obvious what is parallel and what is in series...