C=Q/V12 where Q/ V12 is the Charge per Potential Difference between the plates of the capacitor. If you solve for Q, you see that the charge is proportional to this potential difference. You are likely to surpass your load requirements by increasing the charge/discharge amplitude with the 35 volt cap. In other words, your cap will charge up to 35 volts and then discharge that 35 volts onto your load that was set at resonance to operate with 16 volts discharging. Any separation of circuits using this cap would probably fry something on one side or the other over time. Hope this helps.
yellow-violet-brown
Earthing of the negative terminal is a common practice and will not affect the potential across the resistors in the circuit. Calculating this is simple Ohm's Law. For a series circuit: R(total) = R1 + R2 + R3....... Rn In this case: Rt = 120 + 470 = 590 ohms Per ohms law E = I x R. We need to find the current flowing throught the circuit so: I = E/R = 12 / 590 = .0203 amps or 20.3 milliamps The potential across an individual resistor is calculated again by E = I x R So for R1 (120 ohms): E1 = I x R1 = .0203 x 120 = 2.44 V So for R2 (470 ohms): E2 = I x R2 = .0203 x 470 = 9.54 V For a check E = E1 + E2 = 2.44 + 9.54 = 11.98 The .02 difference is due to the use of significant figures.Another AnswerEarthing the negative terminal will have no affect on the potential DIFFERENCE across each resistor. 'Potential' exists at a single point. 'Potential difference' exists between two points.
If there’s enough room on the capacitor, most manufacturers of electronics components print the capacitance directly on the capacitor along with other information such as the working voltage and perhaps the tolerance. However, small capacitors don’t have enough room for all that. Many capacitor manufacturers use a shorthand notation to indicate capacitance on small caps.If you have a capacitor that has nothing other than a three-digit number printed on it, the third digit represents the number of zeros to add to the end of the first two digits. The resulting number is the capacitance in pF. For example, 101 represents 100 pF: the digits 10 followed by one additional zero.If there are only two digits listed, the number is simply the capacitance in pF. Thus, the digits 22 indicate a 22 pF capacitor.This shows how some common capacitor values are represented using this notation:Marking Capacitance (pF) Capacitance (ìF)101 100 pF 0.0001 ìF221 220 pF 0.00022 ìF471 470 pF 0.00047 ìF102 1,000 pF 0.001 ìF222 2,200 pF 0.0022 ìF472 4,700 pF 0.0047 ìF103 10,000 pF 0.01 ìF223 22,000 pF 0.022 ìF473 47,000 pF 0.047 ìF104 100,000 pF 0.1 ìF224 220,000 pF 0.22 ìF474 470,000 pF 0.47 ìF105 1,000,000 pF 1 ìF225 2,200,000 pF 2.2 ìF475 4,700,000 pF 4.7 ìFYou may also see a letter printed on the capacitor to indicate the tolerance. You can interpret the tolerance letter as follows:Letter ToleranceA ±0.05 pFB ±0.1 pFC ±0.25 pFD ±0.5 pFE ±0.5%F ±1%G ±2%H ±3%J ±5 %K ±10%L ±15%M ±20%N ±30%P –0%, + 100%S –20%, + 50%W –0%, + 200%X –20%, + 40%Z –20%, + 80%Notice that the tolerances for codes P through Z are a little odd. For codes P and W, the manufacturer promises that the capacitance will be no less than the stated value but may be as much as 100% or 200% over the stated value.For codes S, X, and Z, the actual capacitance may be as much as 20% below the stated value or as much as 50%, 40%, or 80% over the stated value. For example, if the marking is 101P, the actual capacitance is no less than 100 pF but may be as much as 200 pF. If the marking is 101Z, the capacitance is between 80 pF and 180 pF.
Usually you'll need 100ohm resistor for each led. In this kind of setting, you will get the LED intensity nonuniform because of its differences due tolerance value. the best thing is to use LED driver IC. Another benefit of using LED driver is, you can mix red led with other color. Now, if we wanted to connect a red LED with a forward voltage of 1.8V and a current of 20mA directly to the transformer output then use the following equation: R=V/I = 9 / 0.02 = 450 Ohms. You could use 470 ohms from the E12 resistor series.
In a Resister u can find totally FORE colored rings.The First three rings used to calculate the value of the resister while the fourth ring indicates the tolerance. Let i say an example how to calculate the value of a resister. there are 10 colors commonly using in resister.Each color indicates a value. let me say it. Black-0 Brown-1 Red-2 Orange-3 Yellow-4 Green-5 Blue-6 Violet-7 Grey-8 White-9 A residter has 1.Yellow 2.Violet 3.Brown 4.Gold So yellow value is 4,Violet value is 7,Brown value is 1 So 47(Yellow)(Violet)*10^1(Brown) =470 Ohm and gold indicates +/- 5% tolerance. That is min value of that resistor is 446.5 Ohh and max value of that resistor is493.5 Ohm. Note: Most of the resistor have Four rings only. But aso there are resistors available with 5 rings
No, the value is far too small. If it is the capacitor used for the timing, the time/s will be reduced to one tenth of the deisred value.
Ah, what a lovely question. Well, the 200WV on a capacitor means it can handle up to 200 volts, while the 470uF tells us its capacitance, which is 470 microfarads. Isn't it wonderful how these little components have their own language to tell us how they can help in our creative endeavors? Just like a happy little tree in a capacitor forest.
between 12 to 14 volts are good.
"470 meters per second" is a speed, not an acceleration. Assuming you mean "470 meters per second square", that is the same as a change of 470 meters per second every second. Just multiply by the number of seconds to get the change in speed."470 meters per second" is a speed, not an acceleration. Assuming you mean "470 meters per second square", that is the same as a change of 470 meters per second every second. Just multiply by the number of seconds to get the change in speed."470 meters per second" is a speed, not an acceleration. Assuming you mean "470 meters per second square", that is the same as a change of 470 meters per second every second. Just multiply by the number of seconds to get the change in speed."470 meters per second" is a speed, not an acceleration. Assuming you mean "470 meters per second square", that is the same as a change of 470 meters per second every second. Just multiply by the number of seconds to get the change in speed.
Yes, as long as there is room to mount the larger capacitor. A better replacement might be a 470uF 16v capacitor, which should be smaller than the 25v cap. If necessary, insulating tubing can be put on the leads & the cap can be mounted off the board, or at an angle to the board. Remember to observe polarity when connecting the capacitor leads to the board. Note the orientation & markings on the old capacitor before unsoldering it.
23 volts across 470 ohms will dissipate about 1.1 watts. Power equals voltage squared divided by resistance.
It is 470.
30% of 470= 30% * 470= 0.3 * 470= 141
81% of 470= 81% * 470= 0.81 * 470= 380.7
If you're saying that the lift is .470" with 1.5 rockers and you want to know the lift with 1.65 rockers, then .470 divided by 1.5=.313 lift at the cam times 1.65 =.517"
470
Other than the integer 470, the simplest form is 470/1.Other than the integer 470, the simplest form is 470/1.Other than the integer 470, the simplest form is 470/1.Other than the integer 470, the simplest form is 470/1.