33

Use 9 V block and check if the resistor becomes hot. P = U2/R. So your 110 ohms resistor should burn close to 1 watt. (Voltage of new battery will be > 9 V.)


25

Make a bridge. Feed it with low level audio signal and use a speaker as the detector. Headphones would be better than a speaker due higher sensitivity. If R1/R2 = X/R4 the sound vanishes. Then X = R4 * (R1/R2) In Wheatstone's bridge R1 and R2 are a potentiometer, R1/R2 is seen visually from the relative position of the slider. Originally DC voltage and ...


23

These are 1k resistors, both black stripes are closer together than the two brown stripes on the bottom. The end with thicker and further apart stripes is the end with the multiplier stripe and the tolerance stripe. This is the rule with 5-band resistors, but due to how hard it is to tell them apart it is best to mark every resistor set with its value and ...


13

Why is this form so commonly presented? Sometimes, when faced with several parallel resistors it's easier to think in terms of conductance (amps per volt) rather than resistance (volts per amp). So, we convert each resistance (R) to conductance (G) and add them. Then we convert back to resistance by inverting the summed conductance value. Forgetting that ...


13

You are right that we can use Ohm's law. Look at the math: The current in this series circuit is: $$I = \frac{V_1}{R_1+R_2} $$ And the voltage drop across \$R_2\$ is equal to: $$V_{R_2} = I \times R_2 = \frac{V_1}{R_1+R_2} \times R_2 = V_1 \times \frac{R_2}{R_1 + R_2}$$ $$V_{R_1} = I \times R_1 = \frac{V_1}{R_1+R_2} \times R_1 = V_1 \times \frac{...


8

Well, \$\text{R}_3\$, \$\text{R}_4\$, \$\text{R}_5\$ and \$\text{R}_6\$ are in parallel and they are in series with \$\text{R}_1\$ and \$\text{R}_2\$. simulate this circuit – Schematic created using CircuitLab So the total resistance is: $$\text{R}_\text{total}=\text{R}_1+\text{R}_2+\frac{1}{\frac{1}{\text{R}_3}+\frac{1}{\text{R}_4}+\frac{1}{\text{R}...


8

Every engineer should become adept at estimation. In the case of parallel resistors, here's an estimation trick: Search for the smallest value resistance in the parallel string. The final result will be somewhat less than this value. In your example case of parallel 30, 60, 20 10 ohm resistances, the 10 ohm resistor dominates the result. The result will ...


7

You can calculate it using Ohm's law by applying it twice: once to R1+R2 to find the current in the entire circuit, then again to find the voltage across R2. But it's much faster if you know the divider formula, especially if you are trying to calculate for an unknown R1 and R2. Usually you are trying to calculate the resistances required to produce a ...


6

Well, resistance is always measured betwen two terminals so that's nothing special. Redraw your circuit with the left leg in the middle.


5

Just for a definitional answer: Two resistors are in parallel if each terminal of resistor A are connected to the terminals of resistor B Two resistors are in series if only one terminal of resistor A is connected to a terminal of resistor B Just because it's worth saying: it doesn't matter how you draw them, it only matters how you connect them. As for ...


5

It shouldn't be necessary to play funny games with resistors and diodes. Lighting LEDs with a microprocessor output is a normal thing to do, and shouldn't require any kind of overly clever circuitry. Make sure you are setting the common to high, and not simply "high impedance." You want it driven to a high state, not simply floating. You want to use "...


4

I'm guessing the leftmost colour is grey, but I'm wondering what the device is that would have an exotic high-precision low value resistor on an SRPB circuit board. If you can convince yourself the leftmost colour is to be ignored for value, you have a very ordinary 220Ω 5% resistor. Wikipedia says that a fifth band is sometimes used as a ...


4

As long as the diode is not conducting, the voltage VR1 is determined by the voltage divider R1 over R2. Increasing R1 will increase VR1, as soon as VR1 reached the diode forward voltage( approx 0.7V) the diode start conducting and will clip the voltage to Vf. mathematically: value of R1 @ which the voltage will be clipped: R1/(R1+R2) * Vcc = Vf So R1 = (...


3

The simplest option is to use an N-Ch MOSFET to simulate a push button operation. With the source connected to GND, the drain connected to the button, then the gate becomes an electrically controlled push button. This is a classic open-drain output circuit. As a circuit it should work to interface with most circuits that use a push-button-to-ground ...


3

Enough time has gone by. Your schematic is horribly drawn (perhaps by intention of its author.) Re-drawing it helps a great deal: simulate this circuit – Schematic created using CircuitLab Assuming ideal diodes with \$V_\text{FWD}=0\:\text{V}\$ and tentatively ignoring \$D_3\$ and \$D_4\$ for the moment, you can almost immediately see now that \$D_1\$...


3

A decent manufacturer will specify pretty clearly in the datasheet what is meant. For example, here's the relevant table from one vendor I've used: As you expected, the resistor tolerance is the limit on the resistance at 25 C, and the variation over temperature is covered by a separate TCR specification. But remember the resistance can also change due to ...


3

There are varying definitions for the term ideal diode. A first approximation definition for ideal diode is usually something like, "A diode that is ON has zero volts across it from anode to cathode (\$V_D=0\,\mathrm{V}\$, short circuit), and a diode that is OFF (\$V_D \lt 0\,\mathrm{V}\$) has zero amps flowing through it (\$I_D=0\,\mathrm{A}\$, open ...


3

You can build a circuit like simulate this circuit – Schematic created using CircuitLab Where RQ and C1 build an RC circuit. Take a watch and measure how long it takes for the LED to turn on. Compare with other known resistors.


3

Your NMOSFETs achieves their rated \$R_{dson}\$ at 4V but your STM32 output pins only apply 3.3V. Your PMOS gates achieve their rated \$R_{dson}\$ at 10V but you are only applying 5V to it. Your MOSFETs get hot and your motor doesn't move because your MOSFETs are only partially turning on, acting as resistors that heat up a lot, but don't allow enough ...


2

So we conclude that D1 and D2 both are on and this is meaningless. That's somewhat true. At the same time, it won't make any difference to the rest of the circuit if one of them is on and the other is off. If they were real diodes instead of ideal ones, most likely one would be on and the other would be only nearly on. How two wires without resistance ...


2

What you have calculated is the time constant of the R-C network $$\tau = R\cdot C = 3300\cdot3.3\mu = 10.89ms$$ From a voltage source of 3.3V, this is the time for the voltage across the capacitor to reach ~63% ~ 2.2V. The problem is, what is loading the R-C is the B-E of the BJT. Ideal-case (ignoring leakage), your circuit looks like this simulate ...


2

An RC circuit generates a T=RC delay to a threshold of (1-e)/e, or about 0.63 of the original step. If you trigger at a different threshod, you will get a different delay. Your threshold appears to be the 0.7 V Vbe of a silicon BJT. If the battery voltage is much higher than 1.1 V, perhaps 3 or 5 V, then the delay to 0.7 V will be much less than RC.


2

Your diagram is OK - resistor value may need altering. However, what StarCat says is VERY important. Most white LEDS use a deep blue LED plus yellow phosphor - the two combine to produce white. They do NOT have very much red component. Using a red filter will remove most of the blue and yellow light and the red will not be very bright. Using red LEDs will ...


2

You start with the typical voltage drop at the recommended current. 12V = 2.1V * 3LEDs + 20mA * R Solve for R, R = 285 ohms Then see how much current you get at the maximum voltage drop. 12V = 2.6V * 3LEDs + i * 285 Solve for i, i = 14.7 mA This is probably OK, if it was much lower, you might want to reconsider and only put 2 LEDs in series. This is ...


2

Here is what the display circuit should look like (showing 2 of 3 digits to make the schematic readable):- simulate this circuit – Schematic created using CircuitLab Each segment has a resistor to set the LED current, common to all digits. Each digit has a transistor to pull the common Cathodes of that digit down, with a resistor setting the Base ...


2

You yet have to connect Line 29 with Line 30. Then your curcuit represents your formula.


2

Resistors R3, R4, R5 and R6 are connected in parallel. And this circuit is connected in series to R1 and R2.


2

There are plenty of ways to determine these values. I personally like superposition a lot because it is already part of the divide-and-conquer strategy promoted by the fast analytical techniques or FACTs. Here, superposition applies because we obviously deal with a linear circuit. We thus determine the voltage \$V_{th2}\$ across the 4-\$k\Omega\$ resistor ...


2

5 band resistors have an extra (digit) band (3th) before the multiplier band (4th). Although a bit hard to spot (I never tried it myself), as described in this question you can start reading from the thicker band's side. So in your case it should be $$110\ \Omega \ 1\%$$


2

Consider this diagram where \$|R| = |X_L|\$: - The following has to be true: - The voltage across the resistor (red arrow) and the voltage across the inductor (blue arrow) have to spatially add-up to equal the supply voltage (black arrow). That addition is shown by the dotted lines. The red and blue voltage arrows have to be 90 degrees apart because both ...


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