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8

My LC circuit consists of: L= 33,4 uH R= 0.3 Ω C=32,63 nF Not quite true; there is also the 50 ohm impedance of the signal generator parallel loading the LC. You would need to drive this circuit differently i.e. with a current source if you wanted to measure Q factor: - Resonant frequency is at 152.524 kHz and the 3 dB point is at 153.241 kHz. The ...


7

Is it possible to preserve the amplitude No, you can't preserve the amplitude because the fundamental sinewave “inside” a square wave is 27.3% higher in peak magnitude (or \$4/\pi\$ in proper terms). In other words, the fundamental sinewave hiding inside a square wave is not the same amplitude as the square wave: - You could adjust it after filtering it ...


6

From the link provided: IMPORTANT: DO NOT use for building/household 110/220 V ac wiring There is no datasheet either. Every component has a voltage rating beyond which there can be no reliable operation or characteristics . They can breakdown in different ways at higher voltage. Please do not use for safety reasons.


5

First of all consider this picture: - It paints 4 scenarios all of which apply to this question: - Top left shows voltage and current in phase and the multiplication of voltage and current waveforms produces a power waveform. Note that the power waveform has twice the frequency of the voltage and current hence, it's not usually meaningful to represent it ...


4

The term you are looking for is "bench power supply". Note that the current setting is a limit. Some reading on the subject: Does a hobbyist need a bench power supply? and What is a Bench Power Supply?.


4

Because interference is a problem anyway, having a higher transmission frequency makes the problem worse. Magnetic coupling is proportional to the rate of change of magnetic flux hence, eddy current induction into conducting objects increases proportionately with frequency. The formula for induced EMF and flux is: - $$V = N\dfrac{d\Phi}{dt}$$ Where N, for ...


4

We don't see iron losses equal to copper losses in a transformer. Iron losses are due to the input voltage, so are the same (roughly) at no load and full load. In fact, they're slightly lower at full load as the primary voltage drop requires less flux swing. Copper losses are due to the current flowing, so are small at no load and large at full load. As ...


3

Absolutely not, no way, no how. Both the NEC (or your local electrical code, surely derived from NEC) and UL's White Book are very clear on what is safe for electrical work and what is not. NEC 110.2. Equipment must be approved. That means some authority like UL must certify it as safe. Next paragraph, NEC 110.3(B), Equipment must be used according to ...


3

Three things to consider Component tolerances.The values you have selected might not physically be those values. There can easily be +-10% The inductor will have resistance to help dampen out the oscillation The ring-down method requires you to excite the circuit at its resonance frequency. Your L-C has a resonance at 1.6MHz while you have excited it at ~...


3

It would be pretty difficult to make AC motors, whose speed is dependent on frequency, turn at reasonable speed. They would need many, many more poles. 2-pole motors already run at 3600RPM at 60Hz. If your supply was just 1000Hz, even an 8-pole motor would spin at 15,000RPM, and your question is asking about multiple KHz. That's probably the whole reason ...


2

It appears that your signal frequency is 20 Hz. This is too low for the sizes of the input and output capacitors. The input capacitor and the two base bias resistors form a high-pass filter of approximately 480 Hz. The transistor's input impedance changes the actual corner frequency, but it still is way above 20 hZ. A similar situation is happening at ...


2

You need to calculate using energy. For this problem Wh (watt-hour) is the most convenient. Batteries are normally rated in A-h (Amp-Hour), not A/h (amps/hour) as you have specified Output Energy = 30W * 2 hours = 60 Wh Input energy = 3.5 A-h * 12 V = 42 Wh So, you need more than a 3.5 A-h battery. And this doesn't even account for inefficiencies or the ...


2

0,3 Ohm resistance maybe isn't true at higher frequencies due the skin effect. But the connection to the signal generator should be kept very loose to prevent its 50 Ohm to affect the result. Insert a capacitor in series with the signal generator. Try one which has 100 x bigger reactance than the LC parts in the resonant circuit at the interesting frequency ...


2

Yes technically this is possible ... although I would not recommend bringing this to practice without a firm understanding of the basics! You say low current, but depending on who you ask this might be different by orders of magnitude. (ECG specialist vs welding engineer. You get the idea.) Can you elaborate on the current requirement or application? There ...


2

From Alternating Current Machines-Synchronous Machines If you did this with batteries, you would have a big problem (your short circuit concern), but a delta (or wye) takes advantage of the phase shift of 120\$^o\$. $$ \vec {V_A} = 450\ \angle 0 ^o\ V$$ $$ \vec {V_B} = 450\ \angle 120 ^o\ V$$ $$ \vec {V_C} = 450\ \angle -120 ^o\ V$$ If you do vector ...


2

Follow the National Electrical Codes - use NMC or IMC rated for burial with a live, neutral, and ground and wire it accordingly. Bury the NMC cable at least 18" and th IMC at least 6"and put protection above it before closing the trench - a treated lumber 1x4 works well. Codes exist to insure the safe use of electricity and should not be ignored for the ...


2

Is AC in general by-default expressed in RMS values? Yes. An unqualified AC voltage is usually understood to be RMS. Otherwise, it would be specified as peak-to-zero (Vp) or peak-to-peak (Vpp). Is the "120VAC" here in the US actually closer to 170V peak-to-peak? 120VAC mains is about 170V peak-to-ground. It's actually double that, close to 340V peak-to-...


2

Is AC in general by-default expressed in RMS values? Yes. AC voltages are usually in RMS. So when we say the wall outlet is 120VAC, it means 120Vrms, which is 170Vpk or 340Vpk-pk. The reason is because it's more straightforward to do power calculations with RMS. If I tell you the wall voltage is 170Vpk, you have to jump through some extra hoops in power ...


2

I'll provide a different approach than Jan did (he shouldn't have computed "RMS" for power), which may be a little more directly understandable. I get Jan's average power value, though. (I see he just deleted his posted answer.) Assume \$t_1=5\:\mu\text{s}\$, \$t_2=7\:\mu\text{s}\$, and \$t_3=13\:\mu\text{s}\$ (\$t_0=0\:\text{s}\$). We also know these ...


1

Can I model the motor as a single phase motor One phase of a three-phase motor is usually modeled. You are then working with the line to neutral voltage and the phase current. The electrical power to the circuit, the losses and the electrical power converted to mechanical power are each multiplied by three for the complete motor. does a motor circuit ...


1

Converting 240Vac to 12Vdc involves a lot of subsystems. You need an EMI filter, rectifier, and then a step down switching converter (buck, or H-bridge, etc.). Electric blankets just produce heat and you can cost-effectively due that by just putting the incoming AC on a resistive load and avoid power conversion completely and greatly reducing the cost.


1

120V-277VAC are typically rms voltages, so I will assume you want to measure the rms value. All three of your methods should be able to do that with a sufficiently fast ADC. Method 1 is able to capture the entire waveform without needing a bipolar ADC input, so it is suitable for use with ADCs that can only measure positive voltage. It relies on the ...


1

Should I assume a phase shift of 180 degrees? Yes. The LM1875 uses a split supply. You need a transformer with a center-tapped secondary in the range 12-0-12 V to 22-0-22 V. The primary should suit your local mains supply, of course. Does the board manufacturer suggest a suitable current/VA rating for the transformer? If not there is an application note on ...


1

Differential VS common mode reactor: it matters how do you draw currents, since it is a three phase, then all currents you draw have the same direction, in turn this is a common mode choke as you would correctly find in the literature, catalogues, shop items,... AFIK a choke, differential or common mode is not possible to build with 5 limb core. The extra ...


1

Well AC voltages are usually rated in RMS, because if you put 10VAC or 10VDC into a resistor, it dissipates equal amount of power and thus generates equal amount of heat. But it is approximately right that 25 VAC can approximately be converted to 25 VDC. First of all, there is voltage loss at the rectifier diodes, if roughly 1 V per diode, the drop on full ...


1

Iron losses are only equal to copper losses at one operating point, which has the importance of being the point of maximum efficiency. To understand this you have to know what iron and copper losses are, and how they vary with loading. Iron losses are caused by hysteresis and eddy currents in the magnetic core. For a given input voltage and frequency this ...


1

Yes, but there will be interaction between the circuits. For example, if you were to directly connect two capacitors of two tuned circuits in parallel, the result would be the sum of the two capacitances. To tune multiple frequencies requires some isolation mechanism, e.g. resistors between each tuned circuit. It depends on how far apart the frequencies are,...


1

So my question is when we have a transformer (delta-wye) for example 220kV/110kV, so primary voltage is 220kV, but what voltage is it, voltage A to ground(VAG in video) or phase A (VAB)? The default voltage specified for a three phase system is line voltage irrespective of the type of transformer that connects to it. So, if it is specified as 11 kV then ...


1

As inductance falls due to the saturation of the core, with a voltage source supplying the stimulus, the rate of change of current increases proportionately. Given that Faraday’s law of induction is: - $$V = L\dfrac{di}{dt}$$ The induced voltage remains largely the same. This of course assumes that the voltage supply is capable of delivering the high ...


1

Well, let's solve for \$\text{I}_0\left(t\right)\$. The input voltage can be written as follows: $$\text{V}_\text{in}\left(t\right)=\hat{\text{u}}\cos\left(\omega t+\varphi\right)\tag1$$ In the complex notation we can write: $$\underline{\text{V}}_{\space\text{in}}=\hat{\text{u}}\exp\left(\varphi\text{j}\right)\tag2$$ The input impedance is given by: $$\...


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