Capacitive reactance (expressed in ohms) is inversely-proportional to the supply frequency, so it will decrease when the frequency increases. The following equation applies:
XC = 1/(2 pi f C)
where:
XC = capacitive reactance, in ohms
f = frequency, in hertz
C = capacitance, in farads
Dear Friend, we can observe over speed when frequency is increase in alternating current ANSWER: You will see that the time between peaks will shorten up. If rectified the amount of capacitance to reduce ripple will decrease.
A: Think about it if there is no frequency there is no problem. The problems becomes apparent as the frequency increases because adjacent proximity now becomes capacitors to influence the circuitry
Capacitors have an equivalent reactance of 1/jwC (ohms) where w is the angular frequency of the AC signal and C is the capacitance. As the frequency of the signal across the capacitor increases, the capacitor reactance approaches 0 (capacitor acts like a short circuit). As the frequency of the signal across the capacitor decreases, the capacitor reactance approaches infinity (capacitor acts like an open circuit). So, if you have a high frequency signal (like a step input) the capacitor will momentarily act like a short.
A capacitor stores accumulated charges as voltage. The storing and discharging of these charges help regulate voltage in circuits. As the switching frequency of storing or discharging increases the 'frequency dependant resistance' or reactance decreases and appears as a short circuit at infinite frequency. The capacitor appears as a open circuit at zero frequency.
Frequency increases.
As frequency increases the energy of a wave also increases.
coupling capacitors are generally used to couple the the AC component of voltage to the DC component(biased voltage) of the transistor amplifier . As we know that the capacitor itself has some reactance which is variable with the applied frequency Rc=1/wc where w=frequency in radians = 2*pi*f and f= frequency of circuit. and, V=VC+VIN VC= voltage drop on capacitor VIN= resultant voltage available for the transistor for amplification so as, frequency increases reactance decreases drop on C decreases so, voltage available for transistor increases and now you can analyse yourself for the case if frequency decreases
As wavelength increases the frequency decreases.
Frequency and energy decrease as wavelength increases.
The energy increases as the frequency increases.The frequency decreases as the wavelength increases.So, the energy decreases as the wavelength increases.
energy
capacitors