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When you put two 10 µF capacitors in parallel the equivalent capacitance would be?
Wiki User
∙ 10y ago
With capacitors in parallel you can just add the microfarads.
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∙ 10y ago
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What would be the effective capacitance if two capacitors are connected in parallel?
A: the capacitance will increase. in series it will decrease accordingly CPARALLEL = Summation1-N (CN) CSERIES = 1 / Summation1-N (1 / CN)
Why use capacitors with crystal oscillator?
There are two kinds of crystal oscillators. One operates at what is called the "series resonance" of the crystal. This resonance is the frequency at which the (AC) impedance between the pins of the crystal is almost zero. The frequency is independent of how much capacitance happens to be in parallel with the crystal - its inside the oscillator and part of the circuit board, etc. But, even frequency that the oscillator runs at.The other kind of oscillator oscillates at "parallel resonance"of the crystal. At this frequency, the impedance from pin to pin of the crystal is almost infinite. This frequency depends on how much capacitance is connected in parallel with the crystal. This parallel capacitance is called "load capacitance". Generic signal-inverter oscillator is this kind of oscillator.The common oscillator connection is for the crystal to be connected from the inverter output to the input. And, there is a capacitor at each end of the crystal to ground. The NET load capacitance is SERIES equivalent value of those two capacitors.PLUS stray capacitance from the circuit board and the guts of the oscillator. Suppose that the crystal is rated for 22pF load capacitance. The stray capacitance is about 7pF. So, that leave 15pF to be made up from discrete external capacitors. If the external capacitors are equal, then their equivalent is half of their individual value. Thus, in this case, we would want a pair of 30pF capacitors.
What if 10 micro farad capacitor is connected to 5 micro farad capacitor in parallel what would be the result?
Since the total capacitance for capacitors in parallel is the sum of the individual capacitances. I'm sure that you can work it out for yourself!
If a 0.001 and a 0.015 capacitor are placed in parallel. what would be the total capacitance of the circuit?
Let C1=0.05F & C2=0.025F, C1 & C2 are in series Let the equivalent capacitance will be C than 1/C = 1/C1 + 1/C2 1/C = 1/0.05 + 1/0.025 = 20 + 40 = 60F hence C= 1/60 = 0.0166667 F
You are given a number of 2μfarad capacitors each with a max working potential difference of 10V how would you construct capacitor of 2μfarad capacitance suitable for use upto 20V explain?
As far as potential difference is concerned if they are in series we have to add them. If they are in parallel the potential difference remains the same. So we need two parallel sets to be connected in series. Now as we connect in parallel the effective capacitance would be got by adding the capacitance. If capacitors of equal value are connected in series, then its value will be reduced to half of the individual. So let us construct two in parallel. So effective will be 4 uF but with 10 V Another set of two in parallel has 4 uF with 10 V. Now these two sets are connected in series. Hence the effective would become 2 uF but the voltage will be 10+10 ie 20 V. Hence the problem is solved. So we need two sets in series
You are given a number of 2μfarad capacitors each with a max working potential of 10v how would you construct capacitors of 1μfarad capacitance suitable for use upto 20v?
Connect two units in series.
What would happen if more capacitors were added to a series circuit already containing two capacitors?
A: For one thing the total capacitance will decrease . If the voltage rating are different then more problem will become evident. That is if they are added in series.
What voltage must be applied to fully charge capacitors in series?
To fully charge capacitors in series, you would want to make sure they are all rated for the same voltage, and then apply the sum total of the rated voltage (if they are 25 volt caps, and you have three, then apply 75 volts). It may be worth noting here that this really is an academic exercise, since putting capacitors in series results in lower capacitance. Most often capacitors will be paralleled so a higher capacitance is attained.
Why voltage gain decreases at high frequency in case of a transformer coupled amplifier?
Depends on the elements used in the circuit.1.At Low frequency: The coupling capacitors are used to isolate the AC input and output from DC bias conditions for active devices. These capacitors with the input and output impedance of the active device act as a high pass RC filter, hence the gain falls.2. At High Frequency: The frequency is high, but not as high as the microwave frequencies. There are two reasonsa>The capacitance of connecting wires are connected in parallel the i/p and o/p. When a capacitor is connected in parallel it acts as low pass filter, hence the voltage gain falls. This is when the frequency is high but not high as microwave frequencies.b> The parasitic capacitance's of the active device are connected in parallel with the i/p and o/p terminals. They along with the device impedances act as low pass filter.
How is total capacitive reactance for parallel capacitors using their reactances is calculated?
Total capacitance for parallel capacitors is simply the sum of all capacitor's individual capacitances. This would apply within (reasonably) any frequency, ignoring non-ideal resistance and inductance, so the same can be said for capacitive reactance.
How do you connect a trimmer capacitor?
Trimmer capacitors are used to calibrate (trim) the capacitance in a circuit. They are connected in parallel with some other capacitor, and the circuit is calibrated by observation of response to known stimuli. Since trimmer capacitors are often used in RF circuits, it is generally not possible to make measurement of the capacitors out of circuit because parasitic capacitance at RF is not negligible. Additional question received on April 6th: "I have a 3 terminal trim cap; 2 terminals read continuity, the other not. My question is how do I connect this in a circuit?" In all probability, the two terminals that read continuity are the same terminal, and are simply provided to give three point mechanical stability when soldered into a circuit. You would connect the two terminals that do not have continuity, using either of the terminals that do have continuity, across the capacitor that needed to be "trimmed".
Advantages of using electrolytic capacitors?
the have large capacitance value, power rating is high , and its constriction. To smooth power supply ripples.The output of a sinewave with the bottom waves cut off is a power supply without smoothing.You would get a hum in an amplifier if it wasn't smoothed by a large capacitance,say 400 Microfarads.On a 12volt supply as in a portable tv,the power supply smoother is around 2000 microfarads.The output waveform should be nearly a straight horizontal line with no ripples on it.
