# Tag Info

## New answers tagged circuit

3

Since resistance is defined as: $$R= \frac{E}{I}$$ In the case you've mentioned, by citing a particular forward voltage (Vf) dropped across an LED with a particular forward current (If) through it, we have $$R= \frac{Vf}{If} = \frac{2V}{0.0075A} =267 \text{ ohms}$$ However, since an LED's Vf will stay fairly constant while its If varies since the LED ...

0

A better way to look at the resistance of the LED is to look at the total current and voltage for the circuit. R = V / I = 9 / .0075 = 1200 ohms Since the resistor is 1000 ohms, the LED must be providing 200 ohms of resistance. As mentioned though, this value is not constant like a normal resistor. Also, assuming the battery is actually putting out 9 V, ...

3

The TL07x op amps are not rated for rail-to-rail operation. In connecting pin 4 to ground, you are forcing the inputs much closer to V- than they can support. The data sheet specifies limits on the common-mode voltage input (which means the input voltage in this case) as 4 volts above V- to 4 volts below V+, which means that it won't work as you've connected ...

2

At the very least, you need to bias the + input of your opamp to the virtual ground you want to use. Use a resistor divider between your 9v rail and ground to get 4.5V. This will bias the input to the midpoint of your rails. In this configuration, you will need two separate biasing dividers for the two op-amps. A more rational approach might be to ...

10

Diodes are not ohmic. Resistors have a linear relationship between current and voltage. That is the relationship described by Ohm's law which you used in your question. Diodes look like this: We normally model them as a simple voltage drop when they are forward biased and an open circuit when they are reverse biased. If more precision is desired, we ...

9

You should not consider LEDs (and other diodes) as having resistance, as they do not obey Ohm's Law. The voltage across a lit LED varies only slightly as the current varies. It is best to say that an LED has a (nearly) fixed voltage drop, which varies with the colour of the LED.

0

This is a simple, and simply terrible, charge circuit. But I'll try to deal with it. I'll assume you're trying to deal with a 12-volt (6-cell) lead-acid battery. Also, your question asked when the CHARGING light goes out. For this circuit, never. As long as power is on and the regulator is more than a few volts the LED is on. So it's actually false to ...

0

A. With no power input and a battery voltage greater than (11.0V + 0.6 Vbe + 1.? Vled), current will flow through D6 and the base of Q1 to light the red LED. B. With power connected and a charged battery the same current path will turn on Q1 and light the red LED. The output of the LM317 is about 1.25V + Q1 Vsat greater than the Vf of the red LED and so ...

1

You need to describe much more clearly what you are doing - what do you mean by transformer enclosure? A shorted turn in a transformer radically affects its operation far more than just reducing the voltage because of the reduced number of turns. The shorted turn acts as a shorted secondary so it will pass a high current in a normal transformer and cause ...

4

The best scheme for adding ESD protection to inputs to a digital circuit will depend to large degree on the bandwidth of the signal input. The simplest ESD filering is done with an RC low pass filter. A series resistor in the signal line and then a capacitor to GND. This limits how fast of real intended signal can enter the system however and it changes ...

0

First, see that in the upper wire you have $$I_1=-2 Vx$$ and you can write: $$I_1 = I_2 + I_3$$, where I_2 is the current in the 2nd wire (from left to right) and I_3 is the current in the other one. Now, you have to "decide" where the Ground is and just solve equations. Note that V_Z has the same voltage that the - of V_x. Using that, I got: $$I_2 = ... 1 I can choose the reference polarity of my resistor however I want, for example: simulate this circuit – Schematic created using CircuitLab Now if I find $V_{R1}>0$ that means that the left side is at higher potential than the right side. If I find $V_{R1}<0$ it means the right side is at higher potential than the left side. Similarly ... 0 you've already obtained one equation relating v(a) and v(b), the other relation between v(a) and v(b) is to be obtained by using ohms law as you already have the knowledge of the current flowing and the resistance between nodes a and b. this way you will have 2 equations 2 variables and you can solve it. for future problem solving you should avoid writing ... 0 You can arbitrarily choose direction, at the end of the analysis the result will tell you if your assumption of the current direction was valid(result is positive) or the current was flowing in the opposite direction (result is negative). However, there is a convention called Passive Sign Convention which is nothing more than a general guideline on how to ... 0 Like the users before me pointed out, you can solve it just by redrawing the circuit, and noticing a balanced wheatstone bridge. -1 The problem is that using standard series parallel transformations does not work in this case. You have 3 options. Kirchhoff's current rule, Kirchhoff's voltage rule, Y/Delata transform. https://en.wikipedia.org/wiki/Y-%CE%94_transform https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws 2 Because the inverting (-) input is at 0V (a virtual ground), the voltage across R1 is:$$V_{in} - 0 = V_{in}$$By the same reasoning, the voltage across R2 is:$$V_{out} - 0 = V_{out}$$The thing that might not be so obvious is that if $V_{in}$ is positive, $V_{out}$ will be negative. In that case the current flows straight from $V_{in}$ to ... 0 The circuit is not a Cascode. A normal cascode would be a common Gate stage stacked on top of a common Source stage. This is done to get high voltage gain from the common Gate (top) stage while eliminating the Miller multiplication of $C_{\text{gd}}$ of the top FET, and reducing the Miller effect of the lower FET $C_{\text{gd}}$ by the voltage gain of ... 0 This is similar to Question 175489 It may require compensation of the negative feedback loop. It may also reflect on your construction method - with such low resistances you have to be very careful about the resistance and inductance of the wiring, especially the ground path from the sense resistor back to the opamp. Please show a better schematic with ... -1 Yes, when in parallel combination, 1/(Znet) = 1/Z1 + 1/Z2 + 1/Z3 ..... 3 Let's just consider the first period of the sawtooth waveform, the output pattern will be similar for the later states.$$U_e = 6t - 30, \quad 0 < t < 10\;ms $$Current $I$ flows only when diode is forward biased i.e $U_{QR} > U_d$ Considering $U_p$ as ground reference potential, while the diode is off $U_p = U_r = 0$ For the current ... 0 OK that's nicer. (I'm going to call all your U's, V's) Let's see, first what is the output if all that stuff in the middle is gone. Just Ve and R1.. what's Va? Second question, How much current is flowing through the middle bit, (diode and R2) if the voltage at the output node (Va) is greater than 10 V? Now what happens when the output voltage falls ... 1 Perhaps I have missed something, you are not 100% consistent in your naming of things. IE in the hand written drawing you have the center node labeled "1" but in the schematic it is labeled "2". Anyways, the answer I get agrees with the given answers. I solved it using 4 current loops. Your mistake is when you multiply equation (i) by 3, you do not multiply ... 0 The time domain is pretty easy. $I_0 sin(wt)$ The signal varies with time $t$ with the frequency $w$. The problem with this notation is that i's difficult to deal with delays and comparing the original signal to the delayed one $I_0 sin(wt - T)$ for example. The solution is to transform the signal to the phasor domain. This gives the ability to ... -2 I think its an optical illusion and that's why it looks odd. the 4 ohm and 12 ohm resistors are whats important here. You connect the RL at that junction, so all that matters is that you get the 4 + 12 = 16, and 32/16 = 2A, confirmed by the 2A symbol there. then its whatever voltage ratio you get. if E = IR then voltage is 2 amps x 12 ohms = 24v at your ... 2 You have very probably destroyed the drive transistor (the one with the collector to the bottom side of the relay coil). As the is no Freewheeling diode, the first occurrence of the switch turning off will produce a high voltage spike at the collector (given by v = -Ldi/dt) which, under most conditions, will be a high enough voltage to cause catastrophic ... 1 You have correctly identified which label corresponds to which pin in the schematic. 0 To add a little bit to the accepted answer: in real life, the voltage after the 1 ohm resistor will not equal the voltage before it for only a split second during the charging of the circuit, when you first turn on the 32V source, or for the split second just after you ground the 32V source. The reason is because during these short transient times, "a" is ... 18 You're trying to find the Thevenin equivalent voltage of the network. This is the voltage that the network produces when the output is open. When the output is open, no current flows out or in to the a terminal. Therefore no current flows through the 1 ohm resistor. Since no current is flowing through the 1 ohm resistor, the voltage across it is 0, by ... -1 The general answer is to draw Karnaugh maps and see what you get. In this case, A and B would be along one axis with C and D along the other, and one map for each of the three output bits. When things don't simplify well, then the logic starts looking more like a lookup table. This kind of small function is easily realized by a small memory. It would ... 0 I don't have the Charles Petzold 'Code' book. Reading your question, it doesn't seem tht the book has communicated the basic concept of an Arithmetic Logic Unit (aka ALU). If you search the web for Arithmetic Logic Unit, or ALU, you will find many usable explanations, for example wikipedia Arithmetic_logic_unit All Central Processing Units (CPUs) have ... 0 I don't have this book. The "decoder" transforms opcodes into signals for driving the various execution units (alu, branch, ...) and selecting the registers. Tranditional RISC CPUs have relatively simple encodings where there are almost fixed fields for selecting the operations. This is far less true in CPUs like x86. (Example, some MIPS instructions : ... 1 I know the program the original poster is referring to since I also had played with it and still have an old copy of it with me. It was called "Virtual Labs Electricity". But it has not been in the market for some years, as far as I can tell. It mostly dealt with components like switches, bulbs, batteries, resistors, and fuse. A more recent program with a ... 4 First, let's be clear that "jumper cables" refers to little wires used to connect things on a breadboard, not to connect two car batteries together. The usual answer is that the breadboard provides the multiple connections to a net. The common solderless breadboards with lots of little sockets have 5 sockets wired together in a strip. The provides enough ... 1 You could remove a piece of the insulation in the middle of a cable. Cut another one in two and solder the side you need to the point where you removed the insulation. Take a fitting piece of heat-shrink tubing, put it around your soldered piece, apply some heat and you are done with a branch element. If you don't have the equipment to solder, there are ... 0 Ideally, you would use a charger that independently charges each cell (the Lacrosse BC series chargers comes to mind). If you want to balance the voltages of multiple charged cells, you could easily place them in parallel with each other. Ideally, they will all have roughly the same V & charge before you put them in parallel with each other. Ideally, ... 2 These are tantalum capacitors. I'm not positive, but I would assume that these are 1uF and 10uF, both with a 25V rating. Tantalums are polarized, hence the "++" one one leg. Oops, looks like I left the same info as @BrianDrummond's comment. Sorry Brian! If you post it as an answer then I'll delete this one. 0 Direct answer: no, you do not need a 50V supply. The voltage supply usually is selected according to the highest required voltage of a single component. Lets say you have 1 motor that needs 5V and one LED that needs 2V. You need to select a 5V supply and take care that all other elements that need less will see a lower voltage. This is accomplished by a ... 2 What you are describing is one of the fundamental laws of electronics, called series and parallel circuits. There are many references for such topics, however Ibiblio has been hosting Tony Kuphaldt's excellent series, Lessons In Electric Circuits for nearly a decade. The first volume, 1-DC, will explain how this works and why. The short answer is, you could ... 0 In the very unlikely circumstance that you intend to connect all the motors in series, so they will run simultaneously, you would theoretically need a 50 volt power supply. However, this is unlikely to work well, as the current required by each motor will depend on the load it is required to move. However, you will more likely want the possibility of ... 0 The engineering way to solve this isn't to dive into exponential equations but approximate the current through the circuit as the reverse (leakage) current through the reverse-biased diode D2; that's because our intuition tells us it will be the element that will "oppose the most" the flow of current (give that the resistor is only 1k). The diode, being an ... 0 It is possible that the original transformer has a double primary (0-110,0-110) currently wired in parallel, and that these only need to be changed to be wired in series instead. However, you also need to check out whether it is the only thing running on the mains supply, as others have mentioned. 0 Why don't you just buy a 230V to 110V transformer like one of these: - Then you don't need to dismantle the MIG welder and try and fit a new transformer. It was probably converted to 110V in the first place so it could be used outdoors on a regulation 110V "safe" isolated supply. If you convert back to 230V then you limit the places it can be used in the ... 1 If the intended output of the transformer is 24VAC, you can put any transformer in that gives the same 24VAC output at your desired input voltage. You can even make it a 12VAC in to 24VAC out, or 2kV in and 24V out, as long as the transformer is made for all voltages, properly insulated and has the right power (wattage). Don't go putting in a 5VA ... 3 I'd use these equations:$$10 \mathrm{V}-I\left(1\mathrm{k\Omega}\right)-V_{D1}-V_{D2}=0I=I_s\exp\left(\frac{V_{D1}}{V_{th}}-1\right)I=-I_s\exp\left(\frac{-V_{D2}}{V_{th}}-1\right) (Notice that $V_{D2}$ appears with negative sign in the 3rd equation because D2 is oriented so that it will be reverse biased) You could work out the algebra on ...

0

There is no MAXIMUM power. There is only one power, and it is always the same. There are several ways to solve, including the Kirchoff Laws, though they all become the same, in the end. The answer, above, shows you how to solve for the one power, which is the maximum power, the minimum power, the average power, and "the power".

1

Giving some hints since its been hours... Continuing what @efox29 was saying, you could write two current loop equations (KCL). Current one ($i_{1}$) going clockwise around the left loop and current two ($i_2$) going clockwise around the left loop, yields the following equations: Left Loop and Right Loop Solve Equation 2 for $i_1$; substitute ...

1

"What is the factor that gives active components the ability to control the direction of flow of current in a circuit." AFAIK what sets active components apart is that they can control the amount of current that flows. The mechanisms that are used vary. Relays use a magnetic field to close a switch. An ancient precursor of the thyristor used inonized ...

0

Passive components can transform energy from one form to another: inductors cant transform electrical energy to magnetic field energy and back, and resistors transform electrical energy to heat energy. It's not clear what you mean by "control the flow of current", but passive components can effect the flow of current in a circuit. For example, Ohm's law ...

0

how come the DC voltage source is disconnected when the circuit is in t<0 position There is no problem with the switch closing at t=0. It's just an exercise in understanding how the already-built-up current in an inductor decays to zero when the energy from the power source is diverted.

Top 50 recent answers are included