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A "voltage reference" is a device that provides a precise voltage to be used as a calibration source, in this example the + input of the op-amp (adjustable in this case using a potentiometer). When a potentiometer is included, it is usually a precision one with 10 or 20 turns, and a test point would be provided on the wiper lead to measure the voltage with ...

-1

The answer is simple: you are making a mistake. When only the dependent source is on you get $V_x=0$. Let's add the voltages on the loop, starting on the low side of $R_1$ and going counter clockwise: $$4i_x-R_2i_x-R_1i_x=0 \rightarrow i_x(4-8-2)=-6i_x=0 \rightarrow i_x=0 \rightarrow V_x=-i_xR_2=0$$ You can actually find networks where you can treat ...

3

Superposition of dependent sources isn't prohibited: Superposition of Dependent Sources is Valid in Circuit Analysis. The author has investigated the presentation of superposition in circuits texts by surveying twenty introductory books on circuit analysis. Fourteen explicitly state that if a dependent source is present, it is never deactivated and must ...

1

This circuit triggers the triac when the voltage across the switch, after going through the RC circuit made from the rheostat and the 100nF capacitor, exceeds the diac trigger voltage. After that, the triac triggers and the voltage drops to a volt or two, so the power dissipated by the rheostat is low. It can have mains voltage across it, but only when it ...

0

Here is a great way to look at a problem where all resistances are equivalent: Intuition: Lets (temporarily) remove the vertical resistors and attempt to analyze the problem. After removing the vertical resistors, lets look at what elements exist at the nodes of the voltage source: Connected between both nodes are 3 branches in parallel Inside of each ...

0

Solution: Place the potentiometer (POT) in series right after the 12V DC source's positive terminal. Explanation: If you look at the circuit to the left of our proposed POT and consider it a black box, the black box has some equivalent impedance. By placing a POT in series with the black box, you have a series connection of one DC source and two impedances. ...

1

You will need a memory element such as a relay to hold the current direction of the motor. simulate this circuit – Schematic created using CircuitLab Here a DPDT relay and two microswitches are used. The motor M1 drives to the "left" when the top terminal is positive and to the right when the top terminal is negative. Suppose the relay is ...

1

Setup so each side can reverse the direction. DPDT switches on each side. simulate this circuit – Schematic created using CircuitLab

1

It's not clear from the wording of the question whether you are being asked to transform the mesh into one having four current sources, or, as is more likely, I think, you are being asked to perform a mesh (current-loop) analysis. That is, analyze the circuit by using the current-loop procedure sketched here: If this is right, then proceed to write ...

0

A DPDT switch is what you want, because if the two microswitches don't act together perfectly, you will have a short circuit and a fire. The DPDT layout is just like you have there. Imagine you glued the two switches together side by side. The DPDT is made just like that.

2

This is a publication style that permits poor photocopies to identify big junction dots from dots of stray carbon on old machines. It resembles the output from TinyCad. Notice it omits some of the useful design info like RefDes ,PN , value, tolerance. I like the big dots too. I used to have HP software like this with rubber band autofit connections that ...

2

$IC_1$ is a configured as a comparator, $IC_2$ is configured as an inverting integrator. The output of $IC_1$, $v_{OUT1}$ is either $V+$ or $V-$. When $v_{OUT1} = V+$, the output of $IC_2$, $v_{OUT2}$ is a decreasing ramp. The voltage at the non-inverting input of $IC_1$ is given by $$v_{+,1} = v_{OUT1}\frac{R_4}{R_4 + R_5} + ... 2 Probably: It's a fixed (approximately) frequency PWM or pulse width ratio modulator with high/low ratio of IC2_out controlled by Vin_AC voltage level. [See below for IC2 defn]. Certain conditions need to be met. See below. Because: lh (=left hand) opamp = IC1 rh opamp = IC2 Set Vin = 0 initially. Assume IC1 output high initially and V_C1 = 0V. IC2 ... 1 I've never tried it, but conductive glue (AKA wire glue) might enable you to make a connection to wires if you can't obtain a suitable connector. This probably wouldn't be reliable without some mechanical arrangement to hold the wires with strain relief. It's also fairly non-reversible. 2 Those are carbon traces, so can't be soldered to. Note the fingers at center top. Those are meant to go into a connector that will mechanically clamp the conductive part against metal contacts in the connector. They could also be meant for a "zebra stripe" connector, or something that is screwed down on the PCB that holds the carbon against pads on the ... 2 Either this question is poorly phrased, or there is additional information related to it. We know that increasing the voltage will increase the current through both the conductor and the semiconductor, therefore the question here is: which of the currents will increase more? We know that the conductor's current increases linearly with voltage, but what ... 1 I see the only difference is that conductor have positive thermoresistance coefficient and semiconductor have negative one. So, unless in thermostabilized environment, you should get more current on semiconductor (a) as they both heated up by current. UPDATE (based on comments) That could be true only with several assumptions, like: Conductor is a typical ... 3 Let there be some linear circuit in a black box with two terminals exposed. We measure the open circuit voltage $V_{OC}$ and the short circuit current $I_{SC}$ at the two terminals. Now, according to Thevenin's theorem, the voltage across the terminals is given by$$V_O = V_{OC} - I_O \frac{V_{OC}}{I_{SC}} = V_{TH} - I_OR_{TH}$$Remember, this result ... 1 It's entirely possible for both the short circuit current and open-circuit voltage to be zero. For example, if the network consists of a single resistor (or probably any linear network with no independent sources). In this case you just need to use a different pair of "test points" to find the equivalent resistance. For example, you can use V=0 (the ... 2 Firstly I'd re-arrange your diagram to make more logical sense: - simulate this circuit – Schematic created using CircuitLab I've placed the transistor at the 0V end of the power rail so it's easier to interface an audio signal to it. I've also put a resistor in series with the base to limit the base current to avoid damage. I've also made the ... 5 An ideal circuit diagram has an associated set of equations. In this particular case, the resistor is irrelevant to the KVL equation which is$$1V = -1V which is nonsense. Just as one can write inconsistent mathematical equations, one can draw inconsistent ideal circuit diagrams (which are simply a different representation of a system of equations). ...

1

If you assumed a resistor was in series with each voltage source the resultant voltage across R1 is zero. The series resistor could be nano ohms or tera ohms and the voltage across R1 would still be zero.

0

This configuration is incorrect. You can not put two different voltages sources in parallel. It violates the Kirchhoff Voltage Law.

1

As I know Battery is a constant voltage source. A battery is not a constant voltage source - it has an effective series resistance (ESR) that means on high current loads the voltage on the terminals drops noticeably. So, if your 9V battery drops to 6.7 volts and the circuit is taking 2 amps then the ESR is 1.15 ohms. If a battery were a perfect ...

3

It's not a short circuit because the resistors restrict the current flow. You can't use just one pole of the battery, because there would not be a "circuit" if the current couldn't flow around. It isn't a transistor, that is a programmable unijunction transistor, and it has an anode, a cathode, and a gate (no base). By the way, contrary to what one ...

6

Yes, this is simple. It can be done passively with two diodes. At these low voltages, they can be Schottky diodes, thereby dropping less voltage and being more efficient. Put a diode from each source to a internal power node. When 12 V is present, the internal power node will be at just under 12 V. The diode to the 7.2 V source is therefore reverse ...

1

The five 2:4 decoders are simply being used to construct a 4:16 decoder. Each output from this decoder represents one of the 16 possible combinations of input values. This part is going to be the same for any 4-input function. The only part you need to "design" is which of these 16 outputs you connect to the big NOR gate, and this is simply all of the ...

2

As drawn R4 and R3 are in series making the equivalent charging resistor $R = \frac{1.62 \cdot 1.78}{1.62 + 1.78} + 1.82 + 2.43 = 5.098 k\Omega$ and capacitor $C = \frac{1.4 \cdot 2.2}{1.4 + 2.2} + 1.2 = 2.055 \mu F$ The time constant is therefore $\tau = C \cdot R = 10.47 \text{ ms}$ As pointed out in comments you have neglected R4. If R4 ...

2

Brian Drummond's answer is correct and complete, I'll just add some observations-- that kind of current and duration is common in resistance welding such as spot and capacitive discharge welding. Usually there is a maximum resistance spec that must be met, such as 0.001 ohms. You can calculate that for a wire from the diameter and resistivity of copper at a ...

4

Set up a little spreadsheet and play with some numbers. Given a length of copper, its cross sectional area, and resistivity, you can calculate its resistance. Now there are two aspects of the cable's performance to consider: Voltage drop, calculated from V=IR. If the voltage drop is excessive, increase the conductor's area. Temperature rise. Given ...

7

Intercontinental symbols for fixed capacitor from http://sound.westhost.com and from http://www.learnabout-electronics.org and a very old one....

1

Simplest solution would be to buy a car battery charger that brings 220V down to 12V attach it to a 12V lead acid battery and wire that in parallel to a 12V pump. No switch, no inverter, and as soon as the power goes out, your pump continues running. The only caveat here may be, depending upon the design of the charger, you may need to add a diode to make ...

2

The simplest approach is to plug the 220V pump into a small UPS. They start at about \$40. There are three things required beyond the pump you have now- a backup power source such as a 12V battery, some means of keeping the battery charged, and a power transfer circuit. The UPS supplies all three plus an inverter so you don't need to purchase another (12V) ...

4

The simplest approach is to get a SPDT relay whose coil matches your primary power source (220 VAC), and whose contacts are rated for your secondary power source (12VDC at whatever amps). Use the normally-closed contact to switch the secondary power to the secondary pump. Whenever the primary power is available, the relay will pull in and prevent the ...

0

I had the same doubt, so I cooked up the following derivation after looking up various links such as those given by Andy aka. My derivation of the formula gets every term in it except the factor of $\sqrt{1-\mu^2}$ which is puzzling me. Here goes - For the first inequality, note that the time required for a capacitor to discharge to $99\%$ of its ...

2

Your reference oscillator p-p voltage is a 5V logic level: - The chip is a standard 4000 series CMOS part that has voltage levels of: - You're running at 10V and the guaranteed minimum high voltage level on a 10V supply is 7.5 volts. Do you see where this might be your problem?

1

A simple way to do this, especially if you're switching 120V, would be to use a relay. Look for one that is either normally-closed ("NC") or has Form-C contacts. Form-C combines NO and NC contacts. Sometimes, in relays, Form-C is called "CO" or "SPDT": You'll need some power source for the relay coil. Since this power is getting interrupted by your ...

0

Does the schematic symbol give you any ideas? When you push it down, it breaks the connection. There are a zillion types of switches, I doubt very much that you need to re-invent the wheel (or N.C. switch in this case). Edit: If I have misunderstood and your intent is simply to reverse the action of a N.O. switch to normally closed, one method is to ...

5

Procedure: Identify all the prime numbers between 0 and 15 (2, 3, 5, 7, 11, 13). Determine how many bits you need for a maximum value of 15 (four bits). Construct a Karnaugh map of the appropriate size and mark all prime numbers as logical 1 and all non-primes as logical 0. Reduce the Karnaugh map to find your logic function. I'll leave you to do the ...

-1

Excess 3 code is just "BCD + 3" code. so you can directly compare your requirement with BCD to 7 segment code available in the internet. This shows one.

2

A loop of wire, with the ends not touching nor connected to anything else, is this: simulate this circuit – Schematic created using CircuitLab If you look at the schematic, it even makes intuitive sense, graphically. As you can see at the gap between the capacitor "plates", the "wires" are not touching. That the plates (which are really just ...

4

Consider electricity as having two co-existing components (this is not strictly correct but it's good for beginners). The first component is "voltage". Voltage is an electric field which propagates at a significant fraction of the speed of light through a conductor (I think roughly 200 million meters per second). When you first connect the red wire to the ...

6

This question is about steady-state dc analysis, and my answer applies to the steady state condition. Is this even true? why wouldn't some of the electrons from the battery on the right still try to flow to the battery on the left? (this kind of relates to question 3). The battery on the right is not part of a complete circuit. Therefore no current ...

4

This kind of question is difficult to answer because you are including too many myths in your question that need to be dispelled before you are in a position to understand a real answer. In other words, you need a basic introduction to electricity, which is too much material to teach in a answer on this Q&A site. I'll therefore just mention a few ...

2

As pointed out by Gsills, there is no negative power rail for it to go negative. The LM358 is a pretty old, and for general purposes where you would 'expect' rail-to-rail ( 0V -> VCC) outputs it doesn't cut it. Your comment of using an MCP6002 is a good idea, it will definitely get the job done. Just make sure it's slew rate and in general it's bandwidth ...

1

If you wish this circuit to function with only a single 5V supply available you are going to have create a negative rail somehow. A DC-DC converter or a 7660 charge pump will do it. Some use a MAX232 to generate +/-9V with the help of a few capacitors.

3

It needs a negative rail. Without that it can't go below ground. Also the LM358 isn't really rail to rail. It can work near ground, but not at it, and it needs about 1.5V from the positive supply.

4

This (original source is here) with some "explanation". This is a pretty dubious circuit in general and I really doubt an ordinary non-differental photo interruptor will work satisfactorily as a smoke detector at non-lethal smoke densities. If you crank P2 down all the way it will burn out T2. It depends on leakage to keep T4 off. There are other issues. ...

0

redraw this circuit as simulate this circuit – Schematic created using CircuitLab You can ignore resistance R3 from the circuit since there is a parallel non resistance path across it. Circuit is then simplified in to R1 // R2. That is 50 ohm resistance in parallel with 1A current source. Circuit is equivalent to 50V( 1A * 50ohm) with a series ...

5

But in AC, their movement is constrained. They move at 60Hz. This shows a misconception. The AC frequency (60 Hz) tells you how often the current switches direction, not how fast the individual carriers are moving. If I hook up 24V, they move faster so more charge passes a area cross section for i=C/s. This is true to a point. But you have to ...

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