# Tag Info

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OK, I think there's many questions "hidden" in this one. First of all, yes, lower volume means that the amplitude of your speakers vibrations is lower. Why? Because the device with the knob is designed to change the way it operates the speaker when you turn the knob. That can be as directly as "the potentiometer is used as a series resistor, so when that ...

4

Buy a new meter. You have learned a lesson but you will pay for it after the learning.

3

This sounds like a power problem. If you are using TTL chips, be aware that they are more power hungry than the equivalent CMOS chips. It is not necessarily a battery problem. Breadboards are not known for being the best at power delivery, and there are certainly quality differences between breadboards. You're going to have power losses - voltage drops - ...

2

Answering about Ohms law: Ohm's law has sometimes been stated as, "for a conductor in a given state, the electromotive force is proportional to the current produced." That is, that the resistance, the ratio of the applied electromotive force (or voltage) to the current, "does not vary with the current strength ." The qualifier "in a given state" ...

2

This type of heater has a thin resistance wire inside a stainless/inconel/other corrosion resistant metal tube, insulated by a ceramic. These elements normally have some leakage, especially as they age. Measuring the leakage to the water is hard. You'd have to measure the two phase lines, accurately, and subtract to the the difference, or use a doubly ...

2

I did salvage about 1 ft of the wire to test it. The wire is about 1mm thick with some type of thread in it about like a human hair. I connected the wire negative to positive on a 1.2v AA cell, a 3.7v cell and a 7.4v battery pack. All fully charged. No spark. UnNo heated wire. No heated batteries. No ohm reading on the wire itself. No battery discharge at ...

2

The user "rdtsc" in has identified that this "pump" is actually a valve, in a comment. You shouldn't use a resistor with it. The valve is designed to be connected directly to 6V, and at this voltage, it will use about 220mA (assuming those are the correct numbers). The resistance of the valve itself is what you calculated (22.72 ohms). If you put a 22.72 ...

2

That's a solenoid, not a pump. It basically allows or disallows airflow from an external source of pressure, it's like a mechanical on/off switch. If you hear a 'thunk' it's working correctly. To address some of your questions in the comments: People don't use motors with resistors because it's grossly and unnecessarily inefficient. The power you waste ...

2

The problem was that I needed to lower the voltage on the base pin of the BJT so the user suggested I add the voltage divider (R2, R3) That's not really what's going on here, but I guess that's what you're asking about. To use a voltage divider to lower the applied voltage you'd need a resistor in series with the switch, so that when the switch closed it ...

1

I need to regulate it to 1to 4ma maximum output without having much voltage drop... Can I just use a resistor in series? No. A resistor works linearly according to Ohm's Law. Resistance = Voltage / Current, so for a maximum (short circuit) current of 4 mA the resistance needs to be 12 / 0.004 = 3000Ω. However at 1 mA the resistor would drop 3000 * 0....

1

What's the resistance between pin T1 and T2 of BT136 Triac when the gate is activated? You cannot predict current by a linear resistance , but you can using the incremental series resistance Rs, just as you can with any Zener, LED diode, Switching transistor , SCR and Triac. For the Triac the equation with the variables shown in every VI graph are; V_T=...

1

What you might be looking for is this: - On-resistance isn't quoted for triacs because they are highly non-linear when $V_T$ drops to around a volt or so (unlike MOSFETs in their "ohmic" region).

1

Let's try a practical case. Given a strong 18650 fully charged battery, Li Ion Cell ESR ~ 0.05 ohm 3.7V If you had AWG 30 magnet wire coiled up to make a little solenoid and somehow it shorted inside, what would be the resistance , R be to make the hottest wire? Please Cut wire < 1 second, otherwise ...V^2/0.05=273 Watts is explosive. self-heating ...

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Homework - no ready to copy solutions will be written, but some guidance. Connect an ideal voltage source Us to AB and calculate how much current your circuit takes from it. You'll see that the current taken from Us is (QCF) x Us, where (QCF) is a quite complex formula which contains alpha, beta and the resistances R1, R2, R3. This means that your circuit ...

1

I remember having similar confusion learning about analog electronics the first time. The book and my calculator would show some value, and the real-world circuit on a breadboard would be a little different. It's convenient to think of those black lines on a schematic as a perfect conductor, a perfect wire. But in reality, even the wires connecting ...

1

Your hypothetical circuit is invalid...the normal rules of circuit analysis can not be used. The definition of a short circuit is a path that has zero voltage across it regardless of current. Our definition of an ideal voltage source is an element that has some constant voltage (nonzero in your example) across it regardless of the current through it. Our ...

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Would decreasing the wire diameter of nichrome wire to increase the resistance help that much to heat up? This is a tricky question. Temperature rise is proportional to power and inversely proportional to surface area, which is proportional to diameter. So for the same power a thinner wire will get hotter. But (on fixed voltage) power is inversely ...

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Why does increasing R of a RL circuit increase the rate of change of voltage across an inductor with respect to time dVL/dt. The relevant formula is $V= -L\frac{di}{dt}$. This says the voltage across the inductor is always proportional to the rate of current change in it. The minus sign indicates that the back-emf voltage opposes the current change ...

1

The thing which caused the speedup was the increase of voltage, not the increase of resistance. If the resistance were zero ohms the current would grow infinitely with growth rate = voltage divided by inductance. That gives amperes per second if the inductance is in henries. The series resistor gives a limit to for the current. In both cases that limit is ...

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A nice game :-). It appears that you have about zero bypass / power supply capacitors. They are essential. There are many ways that capacitors can be distributed but eg Every breadboard has a substantial electrolytic cap (or equivalent) where the power feed connects. Ideally every IC has a bypass capacitor from ground to Vcc (if two rails) with as short ...

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Note first of all that the resistance of an incandescent light bulb is not constant. The resistance of the filament increases significantly as the filament temperature increases, so the resistance of a cold bulb is much less than the resistance of a hot bulb. Now the first question specifies that the voltage will only change marginally, which is a way of ...

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Conductors like copper and tungsten have a large +ve Tempco (PTC) such that the R.hot=10x R.cold. You know that when voltage is applied across 2 series connected R's , the resistor with the largest value sees the most voltage. 40W Bulbs have 50% higher resistance than 60W bulbs so they see 50% more voltage and thus due to the square law $V^2/R=Pd$ the ...

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