# Tag Info

11

In the first circuit, you have one (single) voltage source and one (single) resistor. This one (single) resistor is connected directly across the voltage source terminals ( terminals $A$ and $B$). Thus, from point $B$ to A the voltage is equal to the battery terminal voltage $V_B$ and because our single resistor is also connected directly ...

9

Your 2 × 100 Ω resistors are in series so your total circuit resistance is 200 Ω and this will restrict the current to half of the value obtained in the single resistor circuit. simulate this circuit – Schematic created using CircuitLab Figure 1. An equivalent circuit using a potentiometer. Here we've replaced the 2 × 100 Ω resistors with a 200 Ω ...

9

D2 generates a stable voltage, while the forward voltage of Q1 varies with temperature (by about -2 mV/deg. C). You can compensate for this by putting a standard diode in series with D2 -- e.g. a 1N4148. Another transistor like Q2 with its base-collector tied together would also work well. Then you need to change R1 as the current is then 2.5 V (from D2) ...

6

I don't think you can make any kind of meaningful measurements of short events while using a multimeter. Multimeters react slowly. The resistance of the coil is of less importance than the impedance. The impedance depends on the rate of change and the inductance of the coil. The rate of change will vary as the magnet moves. Measuring the current as you ...

5

You can use a precision current source like TI REF200, then use current mirror to multiply the reference current. Or you may look for Linear Technology, Analog Devices, Maxim, ST,... their portfolio for current sources and applications. It also depends if your current source has a common ground or common positive, and how precise it should be. See, 1uA ...

5

You need to break the circuit and put the meter in SERIES with the LED+resistor+battery. It will measure the same current wherever you put it, since it's a series circuit. You are putting it in PARALLEL (across the LED) which means the LED is shorted (almost) and the current comes from the battery + resistor so you get (ideally) 90mA = 0.09A not 0.009A with ...

4

With a capacitance of 1 F and a resistance of 1 MΩ, your circuit's time constant is $$\tau = RC = 1\,\mathrm{F}\cdot 1000000\,\mathrm{Ω} = 1000000\,\mathrm{seconds}.$$ You would need to simulate the circuit running for many days to see significant charging or discharging, regardless of applied voltage. Instead, you should use much smaller values for either ...

4

Since you measure 12 V the capacitor is already charged, which is as expected because the simulator tries to find the steady-state solution. The 12 pA shown is likely a leakage current, either artificially introduced by the simulator to make the calculations easier, or a tiny parallel resistance in the capacitor model. To see the charging current you have to ...

4

Neither, if the inputs come from voltage sources. The voltage at resistor output would equal to whichever of the two input voltages is higher, minus the voltage drop of the diode. Of course, if the inputs are not fed with a voltage source, the situation is more complex. If the inputs are driven with current sources, then the currents will add up, but ...

4

A cheap and dirty way is to use a voltage source and place a really high series resistance on the output that dominates and the load resistance even as it changes. That way, even as the load resistance changes, it doesn't really affect the overall resistance seen by the voltage source so the current remains more or less the same. 10V in series with ...

3

Behavior at low currents is not given in this particular datasheet, neither typical nor any guarantees. In part it depends on how you define "breakdown" (at what current). If you simulate a model of BZX84C12L from OnSemi in LTspice, you'll find this "typical" response modeled. Of course we don't know how accurate the model is, especially ...

3

LM317 Discussion The LM317 requires at least $3\:\text{V}$ between its input and its output in order to function. When acting as a current source, it will need another (up to) $1.3\:\text{V}$. See the datasheet entries indicated here: So you must plan to lose at least $4.3\:\text{V}$ of voltage overhead if you use the LM317. If you are supplying $10\:... 3 GBW is useful to characterize the performance of an opamp, because most opamps have a "dominant pole" that causes the open-loop gain to decrease as frequency increases. This frequency response shows as a straight line on a log-log graph, and it can be defined by the frequency at which it crosses unity gain. It turns out that all other points on ... 3 I had a quick look at the datasheet. I'm not familiar with this particular microcontroller, but I've used a lot of other Cortex M0/M3/M4/M7 chips, including low-power and high-speed ones. So I'll answer in slightly more general terms. This microcontroller is described as an "ultra-reliable, general purpose automotive" device. That doesn't sound ... 3 The voltmeter shows indeed 1.2 mV. This value is pretty the same each time I driftly pull the magnet. Now when I do the same experiment but by using an ammmeter in series, I see 0.12 mA. So, if your imperfect ammeter has an input resistance of 10 ohms (it's never zero BTW) then you will see 0.12 mA. Try putting a 1 ohm resistor across the ends of the coil ... 3 Firstly your picture Voltage adder Within constraints it's a pretty decent$\boxed{\text{voltage multiplier}}$but, not a voltage adder unfortunately. So, using two diodes you can get a pretty good linear signal multiplier (modulation): - The raw output looks like this: - The red signal trace is the modulation; a 10 kHz triangle wave and, the blue trace ... 3 Yes, you can buy a multimeter that can measure 6nA. For example, the 7.5 digit 34470A has a 1uA full-scale DC current range, accuracy in the 50pA neighborhood. It's a nice piece of kit, but a bit expensive. For cheap measurement you can use a common handheld multimeter if you know the input resistance. For example, the +/-199.9mV range on a 3.5 digit ... 2 Whats the difference between 1ohm & 10ohm resistors or 1kohm and 10kohm resistors? Consider the voltage at the inverting input$V_{IN-}$when looking at the circuit: - We know that an op-amp has massive open-loop gain hence, for practical reasons (in most op-amp circuits) we can assume that the voltage at$V_{IN-}$equals the voltage at$V_{IN+}$... 2 LF351 op-amp to amplify a 0.5Vpp 50kHz signal to a 25Vpp 50kHz. This can be done over a wide range of 50R, R values. For minimum values, current limiting and max swing reduction begin to have effect at 10k for the load on CMOS and in this case, with a bipolar output, they show specs for 10k and 3k to show some reduction differences. Power consumption for a ... 2 Probably Multisim has a similar setting, but in LTspice you need to check te "Start external DC supply voltages at 0 V" to see the capacitor charging. And then you will be able to see the curves: 2 here you have to apply voltage divider rule to understand voltage drop distribution. here is some reference link: - https://www.electricalclassroom.com/voltage-division-rule-potential-divider-circuit/ In ur 1st case when a load is only 100ohm, the voltage drop across the Resistor is 16V. but in 2nd case when you have two resistors in series, so total ... 2 Being sarcastic is not my habit so, even if very good answers have already been posted, I'll try it too. You seem confused by the fact that in both cases the resistors are the same but not the voltage across them. Mhh..without saying anything about what you do not want to hear (ohm..my god I said it !) R3 is not alone: R4 has its influence. So you cannot ... 2 It is because there is half the current. The amount of voltage dropped by a resistance is directly related to how much current is flowing across it. It is a 1 to 1 relationship. 2 The minimum specified value is 1mA where$r_{dif}\\$ aka "Rs or Zzr" rises at lower current from 10min to 50 min thus the drop of 4mA could be a drop around 125mV from a min of 11.7 for C parts. The rated specified value is 5mA. The maximum value of Iz depends on the power limit of Pmax=250mW @<=25'C with a temp rise of 0.33'C/mW of junction ...

2

Source: https://www.nxp.com/docs/en/data-sheet/S32K-DS.pdf In short the chip will consume power depending on: What peripherals are enabled Current consumption of peripherals (use high impedance if possible) What clock speed chip is running at The clock speed pf the S32L142 is determined by the PLL (and if you select an internal or external oscillator, it ...

2

An intuitive way to look at is that all the voltage is dropped across two resistors, and since the resistors are the same, the voltage drop across each will be the same, each taking half. This is called “symmetry”.

2

I just stumbled across this on a suggested reading list, and read because it seemed odd on my list. Teaching IT I've kind of developed a feel for when students aren't sure how to ask the question they really want to know. You mentioned "intuition" so I think you're looking for analogies to your own actions. Rather than an Ohm's Law question, maybe ...

2

It is simple algebra V=IR or R=V/I or I=V/R. On the left, current is I=V/R=16/10 = 1.6 amps so V=IR=1.6*10=16 volts (drop) For both resistor on the right, current (I)=V/R=16/20=.8 For EACH resistor on the right, voltage drop=IR=10*.8=8 volts.

2

The Widlar Current Source should be mentioned, famous for making small driving currents with modest-value resistors. Not that large resistors are hard to come by, but in this case the advantage conferred is a wide range of acceptable load values (good compliance voltage). That quality could be even more important if you are trying to do this with supply ...

2

Do some more reading of the datasheets. Devices like MOSFETs can be offered in several packages, from large TO-247 style to SMD with a small metal tab that solders to the board. So the package used determines the amount of metal area to dissipate heat. For many popular MOSFETs the change shows up as a modified part number, with the base part number the same. ...

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