125

Just ask her what the voltage across the resistor is


96

Method 1 simulate this circuit – Schematic created using CircuitLab Figure 1. A simple practical experiment. Performing an experiment with the circuit of Figure 1 would demonstrate that parallel voltage sources don't change the current. You should get a reading of 9 mA with either or both batteries in circuit. Method 2 A thought experiment: ...


78

There are a few ways to get 5V from a 12V supply. Each has its advantages and disadvantages, so I've drawn up 5 basic circuits to show their pros and cons. Circuit 1 is a simple series resistor - just like the one "some people" told you about. It works, BUT it only works at one value of load current and it wastes most of the power supplied. If the load ...


78

The worst case won't get any better. The result of your example is still 2 kΩ ±5%. The probability that the result is closer to the middle gets better with multiple resistors, but only if each resistor is random within its range, which includes that it is independent of the others. This is not the case if they are from the same reel, or ...


77

It is a regular sense resistor with a kelvin connection or four terminal sensing. The dots are there to show a connection with the wires. A kelvin connection measures the current through the sense resistor for the DC to DC converter. Lord Kelvin is attributed for being first to use the technique to measure low resistances. It is important to make the ...


73

Here is a quantitative way to determine the boundaries of acceptable gate termination resistance \$R_g\$ for power MOSFETs . This will be a lazy lazy lazy (\$L^3\$) approach. So: Very simple FET model, just \$C_{\text{gd}}\$, \$C_{\text{gs}}\$, and \$R_g\$ included. FET capacitors regarded as linear only. FET gate has been pulled down to the ...


73

Let's try this Wittgenstein's ladder style. First let's consider this: simulate this circuit – Schematic created using CircuitLab We can calculate the current through R1 with Ohm's law: $$ {1\:\mathrm V \over 100\:\Omega} = 10\:\mathrm{mA} $$ We also know that the voltage across R1 is 1V. If we use ground as our reference, then how does 1V at the ...


73

Those resistors aren't doing anything at all. All the pins in a row are connected together.


72

Those are just junction dots like all of the others nearby, to show where 3 wires connect. This is called a Kelvin connection. The idea is that the connection should be as close as possible to the resistor. There are also some 4-terminal resistors made especially for this purpose. See Four-terminal sensing on Wikipedia for more information.


71

Clearly to get 18 ohms, you need some of the resistors in parallel. So, what do you need in parallel with a 30 ohm resistor to get 18? Answer: 45 ohms. Now, you have a simpler problem: how to make 45 ohms from three 30 ohm resistors. That should be obvious! Some of the earlier answers have already "given you a fish," but "teaching you how to fish" is ...


66

You are right in that power is the product of voltage and current. This would indicate any voltage x current combination would be fine, as long as it comes out to the desired power. However, back in the real world we have various realities that get in the way. The biggest problem is that at low voltage, the current needs to be high, and that high current ...


65

I agree with some of the others here... you're trying too hard. As others have mentioned, the forward drop of an LED varies with its bias current, but for almost every application a hobbyist will get in to this isn't something you have to spend a great deal of time worrying about. Almost every handheld multimeter has a diode setting. It will tell you the ...


65

A direct solution can be found through the application of continued fractions. If what you have is 120Ω and what you want is 80Ω, write down the fraction: $$\frac{80\Omega}{120\Omega} = 0.6667$$ Since the integer part is zero, you'll be starting by putting resistors in parallel. Invert the fractional part: $$\frac{1}{0.6667} = 1.5$$ This ...


65

This is a crude current source. The two diodes are to create a voltage drop of approximately 1.4v, which is then applied across the transistor VBE and R21. This gives approximately 0.7v on R21, or 10mA through it. Those two diodes could be replaced by a resistor to give the same voltage drop, but then if the supply voltage varied, the current would vary ...


59

The "K" at the end is the tolerance value. "K" is 10%. The one marked 10ohms J is a 5% tolerance part. Your 10 ohm K can be anywhere from 9 to 11 ohms. The 10 ohm J part can be anywhere from 9.5 to 10.5 ohms. You can find the tolerance codes in this wikipedia article on marking codes. You are planning to replace a lower precision, lower wattage ...


57

It's the way the units work out. Broken down to its form in SI units, a volt is $$\mathrm{V = \frac{kg \cdot m^2}{A \cdot s^3}}$$ where A is amperes. So, when you divide by current to get ohms, you see that $$ \Omega = \mathrm{\frac{kg \cdot m^2}{A^2\cdot s^3}}$$ A farad is: $$ \mathrm{F=\frac{s^4 \cdot A^2}{m^2 \cdot kg}} $$ So when you multiply ...


54

The diode is to provide a safe path for the inductive kickback of the motor. If you try to switch off the current in an inductor suddenly, it will make whatever voltage is necessary to keep the current flowing in the short term. Put another way, the current thru an inductor can never change instantaneously. There will always be some finite slope. The motor ...


54

Power dissipation will be the driver. Using six in parallel allows use of standard resistors which may be a stock item. Using standard parts allows use of automatic assembly equipment. Lower profile. Heat spread out over larger area resulting in lower peak temperatures. Ability to combine to make a non standard value. The 20 Ω in your question is not an E12 ...


49

6.8 volts seems awfully high for a single LED. Are you sure that 6.8 is not the number for all four LEDs? That would make it 1.7 volts per LED, which is more reasonable for a red LED. And that would mean that you are currently pushing 172 milliamps, or almost 3 watts through your resistor. If that is the case, you should lower your power supply to less than ...


49

It's the voltage rating on the resistors that is important here. They are powered from rectified 230 V AC and they need to have the correct voltage rating to suit their application. Two resistors in series having an individual rating of 200 V gives a total voltage rating of 400 volts (near enough if you ignore tolerances on values). Take a look at the good ...


49

Your understanding of how power flows through a circuit needs adjusting. 1. How much power flows through a circuit, and is taken from the battery or power source, depends on how much current flows through that circuit. 2. How much current flows though the circuit is dictated by how conductive the circuit is. If a circuit has a high resistance, it is less ...


48

The short answer is "don't do that." The voltage dropped by a resistor is given by Ohm's Law: V = I R. So if you know exactly how much current your device will draw, you could choose a resistor to drop exactly 7.5 V, and leave 4.5 V for your device, when that current is run through it. But if the current through your device is changing, or if you want to ...


48

Redrawing schematics is a great way to analyze circuits, but also an exercise in why schematics are drawn in particular ways — to more clearly communicate to other engineers. simulate this circuit – Schematic created using CircuitLab The rearrangement above should be a little more clear. If you trace a path from one terminal of the battery to ...


48

Assuming you don't have a 20M range, put two of the identically marked 3.3M resistor in parallel on the 2M range, and you should read about 1.65M. The "1" with the rest of the display blank means that the resistance is too high for that range. For example, if you try to measure a 330\$\Omega\$ resistor on the 200 ohm range you'll see the same thing (with ...


47

While it may be true that distributors don't want to check every single part individually, in this case it is not down to laziness that the 0Ω resistor has a specified rated power of 125mW. As pointed out by @BumsikKim's answer, the datasheet for the series does in fact specify this rating - the distributor product page is correctly representing the ...


45

You don't provide much info regarding the environment that the circuit will be used in or the specific resistor types. If you expect temperature variations (or even temperature change caused by self heating) then temperature coefficient becomes important and the initial measured resistance value may be soon way off. By 10% you probably refer to either ...


45

The idea is that the multiplier replaces the decimal point. This dates back to pre-CAD schematics which were hand drawn and then photocopied and reduced. A decimal point could easily get lost during the copying process. By writing 4k7 rather than 4.7k the risk of these errors was greatly reduced. R was used for a multiplyer of 1 because omega could easily be ...


44

The reason for using color bands on through-hole (axial) resistors is simple -- when they are inserted into the PCB, you can't guarantee their orientation -- there is no top or bottom. So you need a way to mark the value so it can be seen no matter how the part is oriented with the board. Color bands are perfect for this. For this reason, I don't expect ...


44

I'm struggling to understand the exact circumstance where this would be needed. What most folk want with a common emitter amplifier is for the DC collector voltage to be about half the supply voltage. So, when you have an AC signal present at the input, the amplifier's output (collector) doesn't clip the signal asymmetrically. Another way of phrasing it is ...


42

Your meter is affecting the measurement Your voltmeter is also connected to the circuit, and in a different position for each of your two measurements. A "perfect" voltmeter would have an infinite resistance, but any real voltmeter has a non-infinite one. So some current flows through it, and that affects your measurement. So your actual measurements look ...


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