# Tag Info

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It appears that the circuits are given as possible answers to a problem. If this is the case, (A) is not a valid answer. If (B) is correct, you have a DC generator and a DC motor connected in series. If (C) is correct, you have a DC generator acting like a DC motor because it is connected to a battery. If (D) is correct, you have a DC generator connected ...

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You would prefer that a coupling capacitor have very low voltage across it at all frequencies of interest. A cheap electrolytic capacitor might have an ESR of 1-2 ohms which might vary by less than one ohm *(over the audio bandwidth**. The input impedance of a line in might be 10K. There's thus negligible voltage change (-0.002dB) which doesn't change much ...

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For audio applications (as opposed to decoupling) there really aren't any high frequency transients to deal with, so a capacitor with a self-resonant frequency well above the audio range should be sufficient. An electrolytic capacitor is of course polarized, and only useful if the DC+signal is always unipolar. Also, an electrolytic capacitor has a shorter ...

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The simplest way is to use a contactor which has a set of auxiliary contacts. These contacts would be wired as shown in your figure, with an indicator lamp instead of the load. The load would be controlled by the main contacts. Doing it this way, the e-stop only has to handle the contactor coil current, not the load current.

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The purpose of the 4017 IC in this circuit is to "remember" whether the relay should be on or off. If you control the relay using software in a microcontroller (an Arduino, Raspberry Pi, etc), then you can simply store that state in a variable in the program instead. The 4017 IC is used here as a one-bit toggle-able memory and can be replaced by the ...

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Adam Hauns answer is mostly right, but both references say dc generator, ac generators does not have +-, while dc does.

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The left part of the circuit is a voltage comparitor, it compares the two voltages (sensor and reference) and outputs whichever is higher. The reference voltage will be a base voltage that the output of the op amp will not go below. If the sensor input requires a higher than base voltage then it will rise above the reference voltage. This is a common setup ...

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The bottom symbol in B is a fairly common symbol for a motor. The other one seems to be an uncommon symbol for a generator. This page identifies it as a DC generator: http://avstop.com/ac/apgeneral/basiccircuit.html But this page says it's "AC power": http://www.electronic-symbols.com/electric-electronic-symbols/electric-generator-symbols.htm Both pages ...

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DC generator (x 4) and DC motor (x 1).

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Simply stated, the value of the resistor must be selected such that with the desired current through the LED (and the resistor, since they're in series) the voltage across the resistor - Vref - will be 1.25 volts. Then with, say, 20 milliamperes through the LED, using Ohm's law to find the value of R gives us: $$R = \frac{Vref}{I} = \frac{1.25V}{0.02A} = ... 3 The voltage across R will be adjusted to a specific value (about 1.25V for the LM317T) by the regulator. Since the resistance is constant and the voltage across it will be held constant, the current through it will also be constant. The adjust pin takes a very small amount of current (usually less than 100uA), and the rest flows through the load. 0 Thanks Ignacio Vazquez-Abrams and Spehro Pefhany for your advices! So, I have reconnected all wires and seems one relay I got is broken now (I suppose I have done short circuit)... Now, when the signal is ON I have connection between NC and COM. 0 As Spehro said, "The 741 is not a single-supply op-amp." That means that the circuit must use a split supply, not a single supply. +V and -V are relative to ground. You need two power sources. For +V, connect the - side of one source to ground and the + side to +V; for the -V, connect the + side of the other source to ground and the - side to -V. Get rid of ... 1 ORIGINAL - NOT recommended. The original circuit is a lashup design with a number of shortcomings which make it hard to guarantee operation. The suggested alternative below is intentionally 'designed' at the same sort of level but has fewer compromises. (Diode OR gates, "interesting" reset cct). The original is designed to be operated on about 9 Volts. ... 0 First this schematic is not very readable and it's a mess. Try to reorder everything so that is clear and nice, that there are as less line crosses as possible. Second it's not clear what is your supply voltage, a led in simulation won't work in one case - the current is too low. If you have a led with a voltage drop 2V is requiring minimum say 5mA and you ... 0 After some additional research, and simulating the circuit with the parameters I obtained, it seems they are correct. The diode has a very low Is due to the fact its forward voltage is about the open-circuit one (7.56V). So, Is will be much lower than the one from a standard 0.7V diode. For the same reason, when you short-circuit the output, the current ... 3 The calculation for a parallel Resistor R2 to put in series with a resistor R1 to get a desired resistance Rx is: R2 = $\frac{1}{\frac{1}{Rx} - \frac{1}{R1}}$ = -111.11$\Omega$ in this case. Here's an example of how such a result could be realized with an active circuit. First, let's assume that one end of the resistor is grounded (it makes it a lot ... 0 The problem may be with R1 , in your breadboard as it seems it has both its ends in same node. Also just cross check whether you have used correct arduino pin labels , GND , and it seems you are using pin 9 as output for arduino so cross check your code again , make sure it is an output pin and you are sending a HIGH to that pin. Also you should give a ... 1 A uA741 will only be able to get within about 1V of the minus supply typically, and within 5V minimum 2V typical of the + supply with a 2K load. The 741 is not a single-supply op-amp. Increase R2 and R3 to more like 10K each and it should 'typically' give you 10V out with a 12V supply. It's not guaranteed though and is not good design. Better to use a more ... 1 Let's say that you have a 100 ohm resistor with 100 volts across it. Then, by Ohm's law, you'll have:$$ I = \frac{E}{R} = \frac{100V}{100\Omega} = \text{1 ampere} $$of current through it. But, instead of one ampere, let's say you need to take two amperes from the 100 volt supply. Since one 100 ohm resistor takes 1 ampere, if you connect another 100 ... 1 The equivalent resistance of a parallel combination is always less than the lowest resistance present.If you can reach your home from school via two different routes...the importance of any single of them would be less for you (if one is blocked,you can always take the other) than if there was only one route.Now imagine yourself as an electron and imagine ... 5 Mathematically put:$$R_1 || R_2=\frac{R_1 R_2}{R_1 + R_2}$$With your values:$$1000=\frac{100 R_2}{100 + R_2}$$Rearranging:$$ 100000+1000R_2=100R_2$$Or$$9R_2=-1000$$So $R_2=-111.11\Omega$. Negative resistance.Physical? Not for a passive component such as resistor. 5 It is not possible to have that combination. If you put two resistors in parallel, the combined resistance will be smaller. Why? Well it's really quite simple. Imagine pumping water down a pipe. Now add a second pipe, is the amount of water you can get through going to increase or decrease? It will go up, because there is more space for water to flow ... 0 It's probably an array of buttons, but with a confusing looking layout. Where the MCU will compare which row and column is live. Here is a button array available from eBay for an Arduino. You can see that there isn't a pin for every button, only for every row and column. 1 In general you could use superposition** but in this case it's much simpler because the two supplies happen to end up being the same. 15V with 2$\Omega$ in series and 15V with 5$\Omega$ in series, which is equivalent to a single 15V supply with 5$\Omega$ || 2$\Omega$ = 1.43$\Omega$ in series, so the current is 15V/(1.43$\Omega$+7$\Omega$) = ... 4 Calculations which show "show Vin should be 1.02V as no current will be drawn towards negative terminal of opamp" are badly in error. As Peter Bennett, has pointed out, your power supply must be grounded. So the op amp portion would look like simulate this circuit – Schematic created using CircuitLab However, this will also not work. Since the ... 5 For the record, here's one of the circuits you tried: Your problem is connecting the load to the source of the FET rather than the drain. Tie the source directly to ground, and connect the load between 5 V and the drain: simulate this circuit – Schematic created using CircuitLab Now nearly all of the supply voltage can be applied to the ... 1 The source you give in your comment (please edit the question to include that) indicates that 100mA current draw may be sufficient. Hence, you can simply throw away 500mW by putting a 50 ohm resistor (preferably rated at 1W) across the output. Of course it will drain the battery faster, but you should still be able to get a long run time unless the power ... 5 A floating gate has a voltage. Any wire has a voltage. Until you put it in reference to something else, that voltage can be anything. Don't expect it to be 0V or any other known voltage. That gate also has a capacitance. Your body is a capacitance. Every time you touch that gate, you're connecting two capacitors and a charge balance will occur. This will ... 1 It's more of an undergraduate degree sized course than a tutorial, if you're trying to do it from scratch rather than re-purposing existing hardware. If I were going to do it I'd start with an Android tablet, a "digital photo frame", or a Raspberry Pi in a box. Disable the touchscreen and write your own software, using the existing video decode. You haven't ... 0 Current is flowing, but an incandescent bulb glows not because of current, but because of heat. There isn't enough current flowing for the filament to get hot enough for it to emit visible light. 2 Multiplying two input signals is a nonlinear operation, therefore the resulting circuit will be required to employ nonlinear components (mainly transistors of course) in appropriate regions of operation. From the top of my head: Stack two NMOS transistors on top of each other, bias the bottom one in triode region, the upper one in active region. As the drain ... 2 0.6V at 20A indicates a resistance of 0.03Ω. P=V2R, so for 60W you need to supply 1.34V (at 45A). Obviously your transformer is not suitable, however you might be able to modify it to get lower voltage at higher current. Some transformers have a gap between the windings and the core, which you may be able to thread a few turns of thick wire (at least ... 1 In a situation where you are actually dividing one parameter by another that has the same unit, the result in unitless or "dimensionless (i.e. a ratio)" as commented by hexafraction. If your teacher answered otherwise, he/she is either wrong or misunderstood the question. That is the situation if you divide the resistance to common by the total resistance of ... 4 The resulting unit is Ohm. You have the formula up-side-down, but it just happens to give the right answer with 2 Ohm resistors. The correct formula is $R_{eq} = \dfrac{R_1 \cdot R_2}{R_1 + R_2}$, which gives units of $\dfrac{\Omega^2}{\Omega} = \Omega$. 4 The correct equation for two parallel resistances is: \begin{gather} \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \end{gather} Which can be simplified using Algebra to: \begin{gather} R_{eq} = \frac{R_1 R_2}{R_1 + R_2} \end{gather} The numerator has units of $\Omega^2$ (because you multiplied two resistances), then dividing by a resistance gives ... 6 The equation for two resistors in parallel is$$R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2}$$The numerator is a product of resistances so its unit is Ohms squared ($\Omega^2$). The denominator is a sum of resistances so its unit is Ohms ($\Omega$). Dividing the numerator by the denominator gives you Ohms:$$\frac{[\Omega^2]}{[\Omega]} = ...

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That's definitely not an appropriate inductor to use in that circuit. I'm assuming from the options presented in the datasheet you linked that the one you have is part number 00 6122 04 listed on page 2. Its DC resistance is almost 1 ohm and its current rating is only 560mA (and we don't even know if that's average or saturation). For an inductor used in a ...

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It's a common trap for beginners and DCDC converters ! Your inductor is not up to the job. These kind of DCDC converters there is always a LARGE INDUCTOR, it needs to be large so it can handle the current. Well actually it needs to be able to store a large amount of magnetic energy. I did not see such an inductor in your photos ! You small inductor can ...

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I understand that your teacher wants you to use the transistor as a switch to light up the LED.You must add a resistor between the base and the positive supply or the transistor will be destroyed by too much internal heat.Other than that,it can work well if you choose the right resistors(let me know if you need help calculating their values) and LED.

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To give you an idea how these counters are made with JK flip-flops take a look at the 7493 on page 4 in in this datasheet http://www.ti.com/lit/ds/symlink/sn74ls90.pdf.The 7490 is a decade counter while 7493 is just a binary counter. The 7493 has the AND gate (mentioned above) already built-in. If you connect R0(1) to QB and R0(2) to QD the counter will ...

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You are apparently on the right track. Now all you need is a binary counter that resets when it gets to 9. That's a total of 10 discrete steps (0..9) Basic theory should have told you that you need a minimum of 4 flip-lops. 3 flip-flops will count to a maximum of 7 (2^3 -> 0..7), so you need to add the 4th stage to get past 7. The problem is that 4 ...

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Use Ohm's law. Since 9Ohm and 72Ohm resistors are in parallel, the voltage across their terminals has to be equal. And since I3 flows trough the 72Ohm resistor (and no other current), you can calculate the voltage across its terminals using Ohm's law. U=I*R. U=1/3A*72Ohm=24V Or, you can calculate the net current flowing trough the 9Ohm resistor. Because ...

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Yes, assuming you've found all the current values correctly, it's possible to find the voltage drop across the various resistors in the circuit. Vx is the voltage drop across the one resistor, we'll call it Rx. Think back to Ohm's Law: $$V_x = I_xR_x$$ where Ix is the current through the resistor. You've already solved for the various currents in the ...

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simulate this circuit – Schematic created using CircuitLab Trying to power a DC motor from glue logic isn't the best idea. they aren't designed to sink a lot of current. Remember a LED only pulls 20mA or so a DC motor will pull quite a bit more. I would use the glue logic to drive a gate of a Mosfet that would sink the current required to run ...

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This is an easy question if you have a solid knowledge in basics of circuits So These are steps that I will use to solve your problem, First find the equivalent resistance between L and M.And before that you want to ground a particular point of a circuit(this is a must because we want to find the voltage, simply we use that grounded point as the ...

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R1 and R2 are NOT in series. Here is how you can do it: R1, R2 and R3 are in parallel. Calculate the parallel resistance of those three resistors so you can see them as one and call it R123. Now R123 and R4 are in series. They form a voltage divider. Now you should be able to calculate the volatge accross R123 (which is equal to the volatge across R1).

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You can connect DC Motor and LED in parallel with the resistor in series with LED. LED can draw current about 24 mA (4.5 V - 2.1 V)/ (100 Ohms). If resistor is not placed in series with it, The LED will eventually get over heated and gets destroyed. use a fly back diode in parallel with motor if it is unidirectional. If Power is set to 4.5 V, then using a ...

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You should wire things like this: simulate this circuit – Schematic created using CircuitLab You require a current limiting resistor in series with each LED, to limit the current in that LED to 20 mA. The motor will draw whatever current it requires - no need for a current limiting resistor for it, as long as you don't feed it more than its ...

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I don't want to use the delta formula ... That's unfortunate because that's the formula to use. ${{4.28\text{V}-1.38\text{V}}\over{4.91\text{A}-0.67\text{A}}} \approx 684 \text{m}\Omega$ $4.28\text{V}+0.67\text{A}\cdot684\text{m}\Omega\approx4.74 V$ $1.38\text{V}+4.91\text{A}\cdot684\text{m}\Omega\approx4.74 V$

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