I have some questions about multiple voltage sources while figuring out KVL. Ive included a circuit labs circuit which shows my question more clearly.
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Yes, this is a pedagogical question. While answering another recent question, I wanted to refer the OP to concise instructions for using superposition to solve circuits. I found that all the easily found resources online were somewhat deficient. Typically they were unclear about what kinds of circuits superposition applies to, or about the actual method to apply the superposition theorem to a circuit problem. So,   What kinds of circuits can be solved by superposition?  How are different kinds of sources treated when solving by superposition?  What are the steps to solve a circuit using the superposition theorem?
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I have two doubts, requesting you to answer my doubts separately. :)  1) I need a resistance of 'X', so is it better to use a single resistor of 'X' value or multiple resistance of r1+r2+r3='X'? What  I mean is using multiple resistors in series instead of using a single resistor has any advantage? Will it reduce the resistors getting overheated?  2) Consider a 1W 2k2 resistor and a 1/4W 2k2 resistor. Is the heat produced by resistors of different watts different? Which resistor will get more heated in the same conditions (I mean the current, voltage etc given both resistors are same)  Regards, Kiran.
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Why does voltage remains same over parallel circuit. If a resistor is connected in the circuit some of the charge should be transformed into heat and make a lack of charge after the resistor (in my sense). So, what's the reason in it?
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Let's say I have a transformer with 3 coils. When primary coil is connected to 230V, two secondaries output 12V. Wires are color coded, but I don't know what colors stand for and there is no information about that on manufacturer's website.  Is there any relatively easy way to determine which wire is primary, which is secondary and which is start to the secondary and which is end of the secondary coil?
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let me say if I have a pack of lithium batteries connected in series and parallels, and i want to connect it to an Inverter, BUT...BEFORE I connect it to inverter, lets say it gives out 12 volts "hypothetically" some informations like 3.7 volt batteries, and 2500 mAh,  lets say then the pack gives 12 volts in DC and hypothetically again I calculate it ((still in DC)) is .....  W = V* Amp ... then it means wattage = 11.1 V * 5 Ah which is 55.5 W. then we got 55.5 watts, THEN....after we want to change the dc into ac it gives you 120 volts "please hypothetically" just answer me this, ARE the Wattage And Ampere are remains the same as DC? what would it be the result after AC?
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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!
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We have the following diode circuit.We have found when the diode is on and off.Now the teacher gave an extra question.What is the ouput Vo (represented in a graph) when we have an input of  a) \$V_i = 100 sin(wt)\$  b) \$V_i = 150 sin(wt)\$  She then solved this exercise .I see the graphs but I dont understand them? Can you please explain them to me? Here is the solved exercise (Excuse the sloppy drawing)    Here are respectively graph a) and b)
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Im a beginnerish will electronic and im looking for a switch that is only ON if the reference voltage is matched -+ 0.25 volts. So if i have a reference of 2.7v to match and i feed the switch a 2.75v the switch would be ON. But if i feed it  2.5v or 2.85v the switch would is OFF.   This might be a simple google search, but i am lacking the correct terminology.
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How anyone know how to do the question? I know how to reduce it to a window pane of 6/6/6/6 with a 3 in series but i have no idea how to do it.  Well i did manage to reduce the right side of the circuit to a pane of 6/6/6/6 and am unable to determine if its a Wye-Delta transformation (probably not) and I do not know which of the resistors are in parallel and which are in series.  Answer should be 4/3V   (The website which the answer was hosted initially no longer exists)
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If we connect an AC source with a pure inductor when AC source is going through 0 volts we get the following graph of the voltage and current.  There are few things in the graph about which i am confused.  In the first graph/diagram "A" of the picture which shows the voltage when it starts to flow and the corresponding current, the current doesnt seem to be lagging behind the voltage by 90 degrees. What is the reason behind that? Is the statement "current lags behind the voltage in an inductor BY 90 degree" is valid just for the steady state voltage shown in graph "B"?  Second confusion is that for the first half cycle of the voltage in graph "A", current is rising both for increasing and decreasing voltage. How is this possible that current is rising in the inductor for both increasing and decreasing voltage?
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In order to supply more power to a circuit. can you split the power between 2 voltage regulators, in parallel?  Can this overpower one of the Voltage Regulators?
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I am replacing a ballast for a single tube. The new ballast I purchased is designed for two tubes but states that it is also compatible with a single (more powerful) tube corresponding exactly to what I need.  They apparently forgot to draw the wiring diagram for the single tube setup though. The only diagram I have is for two tubes and :               Power Switch    +--------------------------+  Line 1 (H) o----/ ----------|Black    Dual Tube     Red|-----------+  Line 2 (N) o----------------|White      Rapid       Red|--------+  |                        +-----|Yellow     Start      Blue|-----+  |  |                        |  +--|Yellow    Ballast     Blue|--+  |  |  |                        |  |  +-------------+------------+  |  |  |  |                        |  |                |               |  |  |  |                        |  |       Grounded | Reflector     |  |  |  |                        |  |      ----------+----------     |  |  |  |                        |  |    .----------------------.    |  |  |  |                        |  +----|-     Fluorescent    -|----+  |  |  |                        |  |    | )      Tube 1      ( |       |  |  |                        +-------|-       bipin        -|-------+  |  |                        |  |    '----------------------'          |  |                        |  |    .----------------------.          |  |                        |  +----|-     Fluorescent    -|----------+  |                        |       | )      Tube 2      ( |             |                        +-------|-       bipin        -|-------------+                                 '----------------------'   How do I convert this to a single tube?
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Electricity takes the path of least resistance!   Is this statement correct?    If so, why is it the case?  If there are two paths available, and one, for example, has a resistor, why would the current run through the other path only, and not both?
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Note that this is a theoretical question - there is no schematic I can show. I will show some schematic, but it will be a very simplified version of an actual circuit, only for illustration purposes.  Assume I have a voltage converter that takes as an input my main voltage (from a power supply) and outputs a certain voltage, for example 1.8V. It would look something like this:         &ndash; Schematic created using   When connecting my circuit to the P.S, I notice it draws too much current (the P.S shows that).  Since I have multiple voltage converters in my circuit (not shown here), I check resistance between each output of each converter to ground. I see that the resistance between 1.8V to ground is almost 0 Ohms. Now I know the fault is either in the voltage converter or one (or more) of the other components drawing power from that 1.8V.  I desolder the resistor shown in the image to disconnect the converter from the other components and see that the converter is fine, but checking resistance from the point connected to all those components still shows 0 Ohms.  My question is - how would you check which component is the faulty one, without desoldering each suspicious component? As you can see in the image, the 1.8V supply is connected directly to the components, without a resistor/bead.  For the sake of this question, assume I have access to whatever equipment needed (no matter how pricey). I wouldn't want solutions to be limited due to availability of equipment.  Thank you!
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Imagine I have a 10V battery with a capacity of 1Ahrs connected to a 100 Ohm resistor.  I'll get 0.1A (V/R) at a power of 1W (current x voltage) for a total of 10hrs. As a watt is defined as J/s my battery produces 3,600J (1x60x60) in total.  Now, imagine I get a second, identical battery and connect it in series to the first one. I will effectively have a battery of 20V with a capacity of 2Ahrs, all connected to the same 100 Ohm resistor.  I'll get 0.2A at a power of 4W (current x voltage) for a total of 10hrs. This will generate 14,400J (4x60x60).  So, from doubling my energy source I have quadrupled the energy output!  I've obviously made an error somewhere either in my calculations or conceptual understanding, and I would be most grateful to anyone who could set me right.  Thanks!
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I've been stuck on this problem for about an hour.     In an ideal transformer, the same flux passes through all turns of the primary and of the secondary . Show that in this case $M^2=L_1L_2$, where $M$ is the mutual inductance of the coils, and $L_1$, $L_2$ are their individual self-inductances.   My first instinct was the following: If we isolate the response of each coil to the other one, we get the equations $N_1\Phi_1=MI_2$ and $N_2\Phi_2=MI_1$, where $\Phi$ is magnetic flux going through a single turn of the coil and $I$ is current. Likewise, if we isolate the response of each coil to itself, we get the equations $N_1\Phi_1=L_1I_1$ and $N_2\Phi_2=L_2I_2$. I would now naturally want to say "Now just multiply $N_1\Phi_1$ and $N_2\Phi_2$ both ways, and get the desired result!" i.e.,  $$(N_1\Phi_1)(N_2\Phi_2)=(MI_2)(MI_1)=(L_1I_1)(L_2I_2)\implies M^2=L_1L_2$$  But that's incorrect, because the response of the coils to themselves and to the other coil are different quantities (i.e. not necessarily equal).  Realizing this, I write down the more general equations.  $$N_1\Phi_1=MI_2+L_1I_1$$ $$N_2\Phi_2=MI_1+L_2I_2$$  Now even with the hypothesis that $\Phi_1=\Phi_2=\Phi$, I still can't see how the identity $M^2=L_1L_2$ follows. Could you help clear up my confusion?  *Taken from D. Griffiths Introduction to Electrodynamics (4th ed.), Problem 7.58.
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The image of the circuit is shown below and it required to find \$V_0\$,  My first attempt at solving this problem is by changing the current source into a voltage source with 1-V and 2k\$\Omega\$ resistance. The fact that the inverting and non-inverting terminals aren't grounded make this problem look difficult.To the point which, I don't know how to proceed with this question or where to start. I would appreciate any help. Following some thought and another schematic from a hint suggested by Alfred, I produced a schematic representing our work.  And  my solution for the problem is below, Using node equation at nodes A and B we have, $$\frac{V_A-V_B}{1k}=-0.5 \text{mA}$$ $$\frac{V_B-(2+V_A)}{1k}=-0.5-x$$ where x is the current that is sent in the output of the op-amp. Using KCL, at the bottom node near the current source we see that the same current that goes through the op-amp also goes through the \$2\text{k}\Omega\$ resistor. Hence, we have, $$\frac{-V_A}{2k}=x$$ Replacing this in the second equation, $$\frac{V_B-(2+V_A)}{1k}=-0.5+\frac{V_A}{2k}$$ And and solving the equations yields \$V_A=-2 \text{-V}\$ and \$V_B=-1.5 \text{-V}\$