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1

Energy storage in an inductor is in magnetic flux, which is related to the current through the inductor. Therefore, the initial conditions for the circuit are given by the capacitor voltage and the inductor current. If you're only told "the switch has been closed for a long time", then you must determine the steady state of the part of the circuit with the ...


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Being that this is a commercial site you will have to hire in professional licensed electricians to change the wiring that feeds into this panel so that it feeds through a separate electrical fixture with the necessary contactor and control port available. Explain to these professionals what you need and I am sure that they can equip your setup with a safe, ...


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As ammeter measure current so for its accuracy resistance has to be low so we connect small resistance in parallel with galvanometer to reduce its resistance same as for voltmeter as it measure voltage drop whose value increases as resistance increases so we connect resistance in series with galvanometer


2

Here is a little "philosophy" about this clever "trick"... Such elements possess negative differential resistance. They are two types - S-shaped (e.g., a neon bulb) and N-shaped (e.g., a tunnel diode). To make them operate in a bistable mode, you have to drive an S-shaped NDR with a low resistive voltage source (having an almost vertical IV curve)... ... ...


1

I recall people using neon bulbs as bistable elements. The ionization voltage is higher than the sustaining voltage. Basically, the neon bulb is biased somewhere between the ionization and sustaining voltages. A capacitively-coupled pulse is then introduced. A positive-going pulse would increase the voltage above the ionization voltage and the bulb ...


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There are a couple of mistakes in your analysis: Mesh 1 should be 16=2(I1-I3)+V1 Mesh 3 should be V1-V2=10I3-2I1.


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This sounds like a school question. So I'm going to give only hints. 1) what is the RMS voltage? 2) Assuming that the voltage source is perfect (zero source impedance), what does that voltage source look like at the secondary winding of the transformer. In other words, you can easily calculate both the voltage and the effective source impedance given the ...


2

You MUST look at the diagrams in the data sheet and understand what they show. You must read the answers that people give and actually take note of what they say. apalopohapa have you a completely clear answer. You ignored what he said and just asked the question again. He said "The battery goes both to the charger and the circuit." This means The ...


2

The battery goes both to the charger and the circuit. When you are charging the battery, you are still powering the circuit. To turn the device off, you would normally have to either provide a switch (mechanical or solid state) that disconnects the battery from the rest of the circuit , or have the circuit go into a low power mode that won't drain the ...


2

Laplace Transform: Solving in the s-domain will be easier. Writing the current division formula, $$i_c (s) = \frac{R}{R+1/Cs}\times I(s)$$ $$i_c (s) = \frac{1}{1+1/5s}\times \frac{10}{s} =\frac{10}{s+0.2}$$ Taking the Laplace inverse, $$i_c(t) = 10e^{-0.2t}u(t)$$ $$\therefore i_c(0^+) = 10A$$ $$------------------------------------$$ Another view to ...


1

In your final equation, the right hand side is not 0. Just as you said, it is 10delta(t). If you then try to solve the equation (using integrating factor method for example), there is an integration that turns the delta function back to a step function (which is simply a constant for t>0) Now, what would be the initial condition? Let's look at this another ...


1

The diode is for reverse voltage protection, and to prevent the 100uF capacitor from discharging back into the motor rather than supplying power to the control circuitry. The capacitor helps the control circuitry maintain a stable voltage when the motor switches on, which consumes large amounts of current and probably causes the supply voltage to drop. I'd ...


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You can use two or three power-on reset chips with different delay times. Here is one example of what I am talking about: TI LP3470. http://www.ti.com/lit/ds/symlink/lp3470.pdf There are other vendors and other chips with different features. On first application of power, these hold their reset output low, and keep it that way until after some delay time ...


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The placement of the resistor ALWAYS matter at some combination of geometry and frequency. Typically, however, what people are asking is whether they will notice. Just find two old Tandy Radioshack TRS-80s, place them side-by-side and try to run programs on both at once!


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It doesn't really matter where you place a current limiting resistor for LED. I usually place them between power and anode, so I can connect all cathodes together and to the ground when I have bunch of LEDs next to each other. This way I can place the resistors anywhere on the board and will save some space but not having to have traces for return current ...


6

Resistors in series with other components are, themselves, transitive. They can be swapped with other components in the uninterrupted series and they will react the same. However, the other components may not like being swapped. From the resistor's point of view, the current is the same regardless of the position. Since the LED is a current device, it ...


1

If you're asking more broadly than just at the circuit level. For me physical placement of resistors is important for terminating high speed signals on the board. You might use a small resistor at the source to match the output impedance of the driver to the trace, and you might use another at the end to terminate the signal. Same goes for resistors ...


1

In general (that is, almost always) the location of a component in a series connection does not matter. The exceptions generally have to do with the relationship of the part to other components as mediated by considerations that don't show up on schematics. For instance, in high-voltage circuits, a switch in series with a load resistor to ground might be ...


3

It becomes important when there are multiple paths for current to travel in a circuit. A basic led resistor circuit has a single path. If you need to control the path of current, or the value of voltage at a certain node, then the resistor placement (and value) becomes important. As a programming analogy, simple addition of a group of integers, it doesn't ...


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Since you say you've been a programmer for years I am going to ask you to stretch your mind a little bit answering your question in the most general way. You refer to resistors, I am referring to any two terminal component (TTC). What's a two terminal component? You guessed it, it's anything that has two terminals. That is useful since we can define a ...


0

The resistor serves to limit the current that flows through the LED, so it needs to be placed in such a way that the resistor and LED share the same current path. IOW, in series. It does not matter which side of the LED the resistor is on. Also, if you have multiple LEDs in series, you only need one current limiting resistor, and it can be placed anywhere ...


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I can't provide detailed answer of a homework if you don't show an effort as per this site rules. Hints R12=V1/i2 when i1=0. That's it you have a current source i2 at the right and you measure the voltage V1 at the left. The division is R12. Possible way of doing. Steps: 1) push 10 ohms resistor to the right of the transformer. It becomes a 10/4=2.5 ohms ...


1

X10 irfp460 mosfets, you are correct the IR2110 cannot drive X10 irfp460 mosfets directly, local buffering is required, driving mosfets fast requires the gate capacitance channel to charge fast and discharge fast, the higher in frequency you switch it the more current is required. The irfp460 has a Qgs (gate charge) 210nC each, that's 210x 10 mosfets = ...


1

(EDITED) - Notice that the last Nand gate ahead of the beeper looks to be a slow oscillator, (the beeper likely rings as ON-OFF-ON-OFF....). The input to the last Nand changes from being either a low signal or a fast square wave signal, (which corresponds to wet soil then to dry soil conditions). The diode in the feedback of the last Nand actually makes ...


0

Instead of the beeper, find a relay that has a coil compatible with the circuit's power supply and use the coil contacts for activating a motor running from 12V. If the circuit you have shown operates from 12V then it can be easier - use a P channel power mosfet instead of the BJT (Q1). All this assumes that the circuit you have posted does what you want it ...


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A power supply sequencer or power rail sequencer. This could be done with a simple microcontroller for example, or you could use an analog comparator. Essentially you want a circuit which has a transistor or other controllable switch for each power rail. When the first rail in the sequence reaches some threshold (e.g. detected with an op-amp or analog ...


0

The comment link gives some of the basic circuit operations. If that does not provide your answer below is a simple written out explanation for the use of the two resistors. As the name implies, "Programmable Unijuntion Transistor", there needs to be a programmed voltage supplied to the Gate pin. In this circuit the two resistors (15K and 27k) provide ...


0

The 5 Ohm 5 Watt resistor should be a good solution for a simple charger. It will limit the current to about 500 mA ((12.6V - 10V) / 5 Ohm) and dissipate about 1.3 Watts (0.5A * 0.5A * 5 Ohm) which is less than its 5 Watt rating. This will not charge the battery to its capacity, but will not overcharge it either. Adding more 5 Ohm 5W resistors in parallel ...


2

The equation for clockwise should be: $$-v_{s} + (i_1 + i_s)R_2 + i_1R_3 = 0$$ $$-v_s + i_1(R_2+R_3) + i_sR_2 = 0$$ $$i_1 = - \frac{ i_sR_2 - v_s}{R_2+R_3}$$ Which is exactly negative of the other direction choice.


1

You need a pull-up resistor on the output side of the opto, like this: simulate this circuit – Schematic created using CircuitLab If the output transistor's emitter is grounded, it can only pull the collector towards ground, or let it float. To ensure the motor driver receives a "High" when the transistor is off, you need to add the pull-up ...


1

1) Maybe yes, maybe no. The left branch, in particular, assumes an unreasonable precision in specifying the LED voltages. I assume you have access to data sheets for the LEDs you're hoping to use, and that's where you got the numbers. Check them again. Those are probably typicals. If they are maximums, the design is in trouble. Let's say, for instance, that ...


1

Depends on what the LEDs are rated for. The left branch LEDs have a forward drop of 11.8V, and the right branch drops 10.8V. The limiting resistors at the bottom of each branch should restrict the current to 1A, so make sure your LEDs are able to handle that current. A better way to drive this would be with a constant current LED driver. A switching ...


2

You need to connect all the power pins, and should probably do so with both low inductance bypass caps as close to the chip as possible, and larger value supply caps just beyond. If that is one of the parts which requires external caps for its internal regulator (I know the `405 for example does), you will need to provide those between the VCAP pins and ...


0

Since this looks like a homework problem, let me help you with what I know. You have 8 LEDs as output. But since the LEDs are going to be turning on in pairs, what you need would be 4 output lines, each lines to drive two LEDs. Now what you need is four output lines which would be turning on one after the other (only one line turn on at a time). Tabulating ...


0

There are more solutions to this question: You can use uC and program it. It is complex if you are not into programming and that stuff. But requires less components and smaller PCB. You can use timer and counter/divider. It is less complex (you just solder) and requires more components. Both of these solutions are solved over the internet, you need to ...


4

An ideal capacitor has no resistance and therefore no heat will be dissipated by the capacitors in your circuit. The only place in that circuit (assuming all ideal parts) that electrical energy will be converted to heat is the resistor, so what you need to find is the power dissipated by the resistor, which involves the charges stored in the capacitors as ...


1

It was a defective wire, it had nothing to do with the chip.


0

Transistors are not really symmetric. If you turn NPN turned around is still NPN, but that is an over simplification. The emitter typically has a much higher doping level than the collector. You want high doping in the emitter because the higher doping level will give you a higher current gain. It injects more electrons into the base if there are lots of ...


0

Part of the problem here is that the sag of the 6V supply (i.e. 4X AA batteries) does not need to be very much to cause total havoc with the VIN to +5V regulator in the Arduino. For example if you are using the Arduino Uno there is an MC33269D-5.0 LDO regulator on the board to regulate the VIN down to the +5V required for the board operation. This LDO is ...


2

My point of view: measure the time of the power wave, add a Schottky diodeļ¼Œthen add a capacitor to be keep the power when the motors start. Capacitance of the capacitor depends on the wave time.


1

Could you power your arduino from a separate battery from the 6V pack you are using for your motors? Then the arduino could control the motors via a relay or a transistor, and your arduino power supply would not be affected by the voltage sag on the motor circuit. Is the 6V power supply connected to Vin on the arduino? If so, you might be running the ...


3

Yes, to both of your questions. The symbols used for "+5V" or "Vcc" or "ground" are simply notational conveniences which mean "all the stuff with this symbol is connected together". We can also assume, if it isn't explicitly drawn, that there is a 5V power supply connected across "+5V" and "ground". It doesn't mean anything beyond that. It's a common ...


0

If you want to stick with the D-latches you may consider these: Like someone already said you also need to use D-latches with a CLR pin and clear them with a differentiator. 1) You could make use of a RS flip-flop made of OR gates, set it by the Q of the first D-latch and reset it by the Q of the last one. Mind the type of logic you use. 2) You could ...


2

Your logic is, in principle headed roughly in the right direction, but there are a few problems. In the first place the 2-input OR gate is redundant, since the right-most FF ouput will force the NOR output low. More importantly, you show no means to reset the outputs to the correct starting state. Generally the useful approach (using discrete logic) is to ...


1

Use a microcontroller. Something like a PIC10F or 12F would be ideal. For sequencing three signals you can get away with a single 8-pin DIP IC (and a decoupling cap). I'm happy to expand if you're willing to consider going down this path. For anyone that wants to point out that there are SOT23-6 PIC's available, yes, this is true - but for an OP that is ...


0

Capacitor voltage can't change instantly, since that would require infinite current. Therefore the capacitor voltage at T = 0 is whatever it was just before T = 0. At T = ∞, everything is assumed to be in steady state. If the circuit is purely DC, then no current will be flowing thru any capacitor and you can replace all caps with open circuits for ...


0

Voltage across a capacitor cannot change instantenously, so: $$V_0(0^-)=0 \Longrightarrow V_c(0^-)=V_c(0^+)=0 \ \therefore \ V_c(0)=0V $$ You can write the following diff. equation for other values of Vc: $$R3.C\frac {dV_c}{dt}+V_c=V_0$$ Solution of this first order linear diff. equation gives: $$V_c(t)=1+ke^{-t/C}$$ If you use the initial value you ...


0

For a more simplified format (with out the calculus), first find the circuit's time constant RC, which is also known as "tau". Lets use this as "t", so then t=RC. With t in seconds. Once you know t the voltage on C can be more easily calculated. The voltage on C will change by 63% of the applied voltage (applied across RC) after each t time period. ...


0

For a single MOSFET solution you should feed the motor from a +12V supply. Then have the microcontroller generate a PWM signal with varying duty cycles to generate an effective drive to the motor to correspond to the desired 6V, 9V and 12V drive levels. Clearly for the 12V to the motor the PWM would be at or close to 100% duty cycle. For the 6V equivalent ...


0

1) At the instant of turn-on the [discharged] capacitor will look like a short circuit. What's the voltage across a short circuit? 2) When the capacitor is fully charged it'll look like an open circuit. What's the voltage across an open circuit?



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