7

The differential equations which use \$\frac{di}{dt}\$ and \$\frac{dV}{dt}\$ are more fundamental. They do not do not care about any abstractions such as "frequency", "sinusoids", or "canned" waveforms which, in a sense, need you to know what is going to happen in the future. As a result, you can always use the differential ...


6

The answer is "yes". You can't have a change in current without a voltage appearing across the terminals. And if a voltage appears across the terminals there will be a change in current. If you're driving the inductor with an ideal voltage source, it's easiest to think of the voltage causing the current to change. If you're driving the inductor ...


6

It should not be a problem if the parts are small and reasonably close together. 433MHz is a relatively low frequency. The inductance of the traces is important whether there is one inductor or two, only a few mm can have an effect greater than the tolerance of the inductor.


4

Most likely, you can't. There is more to an inductor than just inductance. There is also current rating, winding resistance, core saturation current etc. In power conversion those can be even more important than the inductance itself, because it's where your inductor loses come from. As a rule of thumb, if something seems to good to be true, it is. It's ...


4

What exactly will make the inductance decrease Consider the following tuned circuit that resonates at precisely 800.000 kHz: - If I plot the resonance of Vout whilst changing R1 from 1k to 3k3 and then to 10k, the frequency doesn't move one little bit: - It remains at exactly 800.000 kHz. It gets peakier as resistance increases but that's the Q factor ...


3

My question is, do these formulas always hold? Apart from the physical limit cases when transmission line theory take over, the formulas always hold irrespective of waveform shape: - $$I = C\cdot \dfrac{dv}{dt}$$ $$V = L\cdot\dfrac{di}{dt}$$


3

Note how these inductors are air filled so even though they're physically large in size, that doesn't mean they also have a large inductance. Indeed tapping off a signal is much easier with large inductors and that is I think the actual reason why they're this large. With a smaller inductor tapping will be more difficult, the wires are not isolated, if they ...


3

The analysis clearly shows multiple resonances which are displayed as slight notches because they involve damping. This means that a simple RLC won't do, you'll have to add more RLC cells (there may be reflections, too). This is commonly done. At any rate, judging by the pictures, it looks like either you didn't specify Rpar, or you did but it needs a lower ...


3

A laminated iron transformer is designed to work with low frequencies, usually 50-60Hz and with some derating could go up to about 500Hz. Switching converters, on the other hand, work at usually more than 20kHz, so you saturate the core quite fast. It won't really work, unless probably with extreme derating just to say you made it working


3

Your L1 is a tuned circuit along with C3. If you change the value of L1 you will need to also change the value of C3 to maintain the same frequency. In THEORY a combination of more than one inductor will have the same effect as a single inductor but in practice it may introduce additional parasitics that you will need to accommodate. Often in circuits like ...


3

It could work, but look out for efficiency. Ceramic inductors tends to have quite a lot of DC resistance. If saturation current, average current and inductance are in the range needed for the circuit it can be worth a try.


3

What makes the voltage across the inductor stay constant at Vi? It stays constant because the incoming supply defines that voltage but, if you left the inductor connected across the supply, after a few tens of microseconds, the current might be tens or hundreds of amps and that will cause the incoming supply to collapse. Then you're in trouble. But that isn'...


2

It depends what you mean by 'cause'. If you kick a football at a window, and the window breaks, then you can say the football caused the window to break, and you can say the broken window did not cause the football to move. It's this ability to say that one did not cause the other that establishes an unambiguous cause and effect chain. With inductance, and ...


2

You can technically write KVL for the second loop and find that only \$I_2=0\$ satisfies it. I suppose the purpose of this task is to train you in application of the dot rule (which is simple: mutual inductance e.m.f. has the same sign as self inductance e.m.f. when both the loop current and the coupled current are flowing in or out of dotted ends; the signs ...


2

Your 1st graph has something like pixel aliasing ripple but correct 0.444 A peak Your 2nd has the probe location for 12V-Vc =0 steadystate so not Vc. My sim


1

For a general RLC circuit - 1.Indentify the output asked(given) in question ! 2.Find transfer function i.e $$T(s)=Vout(s)/Vin(s)$$ using KVL, KCL , Mesh , nodal etc. Methods 3.And once you get the Transfer function(which will be 2nd order most likely ) then convert that Transfer function into standard form i.e $$ \mathrm{ H(s)=K \frac{\omega_n^2}{s^2 + (2\...


1

There is no "standard" for this and you will find that even among similar types of appliances the value differs. To find out what type of load your appliances present, you will need to take measurements.


1

My question is, do these formulas always hold? The answer is really "yes" and "no". Other answers have explained the "yes" answer, but they all depend upon a capacitor or inductor being "ideal". Real capacitors and inductors have "stray reactances" and resistances. But even if we ignore these, real ...


1

The impedance formulae always hold true (within specs) but the spectrum of the input signals can be varied from sinusoidal so the response depends on the circuit transfer function. s domain plots or Smith charts or Bode amplitude and phase plots will demonstrate this.


1

I am getting around 80kW/m3 You need to examine the volume of the core used: - So, in cubic metres, the volume of your core is: - $$\dfrac{102000}{1\text{ billion}} = 0.000102 \text{ m}^3$$ So, if the material is capable of working at a power loss level of 80 kW per cubic metre, the actual core set could be used with a loss up to 8.16 watts. Here's an ...


1

The term you are looking for is "gyrator". A gyrator is a passive, linear, lossless, two-port electrical network element proposed in 1948 by Bernard D. H. Tellegen as a hypothetical fifth linear element after the resistor, capacitor, inductor and ideal transformer. ... Circuits that function as gyrators can be built with transistors and op-amps ...


1

Source Those are mine theoretical values. $$V_{rms}=\frac{29.2V}{2\sqrt{2}}=10.32V$$ Peak to peak voltege is divided by 2, so we get a peak voltage then it's divided by sqrt(2) to get RMS value. $$I_{rms}=\frac{10.32V}{\sqrt{574.2^2+5^2}\Omega}\approx 18mA$$ The equivalent capacitance is approx. 1.387nF when you combine both Cs and Cr. So I have inserted the ...


1

Well, the RMS of the input current is given by: $$\overline{\text{I}}_\text{in}=\frac{1}{\sqrt{2}}\cdot\frac{\text{amplitude input voltage}}{\left|\text{input impedance}\right|}=\frac{1}{\sqrt{2}}\cdot\frac{\frac{29.2}{2}}{\left|\underline{\text{Z}}_\text{in}\right|}\tag1$$ Where: $$\underline{\text{Z}}_\text{in}=1+\frac{1}{\text{j}\cdot2\pi\cdot125\cdot1000\...


1

Your description revolves around magnets and mechanical force which makes the term co-energy very relevant. I'm sure you can find plenty of information regarding that. Here is a small example on how magnetic force could be calculated. You have a coil with core as well as an object with relatively high permeability \$\mu_r\$. Then the air gaps are the main ...


1

Your assumptions about \$I_1\$ and \$I_2\$ are not correct. The current through an inductor must be continuous, but these currents are through resistors and there is no inherent requirement that \$i_R(0+)=i_R(0-)\$.


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