Ripple voltage, in the presence of a filter capacitor, is inversely proportional to load resistance.
If the load were zero (resistance infinite), then there would be no ripple voltage. As the load increases (resistance decreases), the ripple voltage increases.
The ripple waveform will appear to be sawtooth, with the rising edge following the input AC from the diode's conductioin cycle, and with the falling edge either being linear or logarithmic, depending on load. If the load is resistive, without a regulator, the falling edge will be logarithmic. If the load is constant current, such as with a regulator, the falling edge will be linear.
The higher the load current the higher the ripple voltage across the filter capacitors. It is directly proportional.
The electric resistance is related to the diameter and extension of the wire submitted to a determined voltage which will determine the electric current flowing into the wire.AnswerVoltage has no effect on resistance. Resistance is determined by the length, cross-sectional area, and resistivity of a material (resistivity is affected by temperature, so temperature indirectly affect resistance).
The readings on an ammeter indicate the current being drawn by a load in a circuit. This load is basically a resistance to current flow. The higher the resistance, the lower the current. The supply voltage has a direct effect on current flow. The higher the voltage applied, the higher the current will be. So the readings will vary on the ammeter according to fluctuations in load and or resistance of the circuit and the applied voltage.
The electrical potential energy increases as the voltage is increased. It further excites the filament in the bulb more than a lessor voltage would. Using good old ohm's law (Voltage = Current x Resistance), a larger voltage applied to a bulb at the same resistance increases the current proportionally and larger currents has the effect to cause higher temps in conductors
There's no effect since the capacitor was already faulty i.e it was like not in the circuit. Install a healthy capacitor because it will improve the power factor of the fluorescent lamp circuit thus reducing energy wasted.
No, it is the capacitors effect upon the motor's winding that starts the fan to turn.
The effective resistance of the capacitor reduces the ripple current through the capacitor making it less effective in its function of smoothing the voltage. But if the capacitor filter is fed by a transformer and diodes, the resistance of the transformer exceeds that of the capacitor.
If the resistance is in series with the capacitor, the charge/discharge time is extended.
The dielectric material between the plates.
A capacitor and a resistor has no effect on the supply voltage; however, this particular load combination will cause the load current to lead the supply voltage by some angle termed the 'phase angle'.
In case of a lossy capacitor, its series equivalent resistance will be large.
No.
it will improve the power factor... The angle between voltage and current will decrease depends on capacitor value.
It will decrease the voltage drop.
By Ohm's Law, current is voltage divided by resistance, so if you double both the voltage and the resistance, the current would remain the same.
A: It is not a bypass it is a negative feedback to effect the input voltage, Bias is when a DC voltage is applied to insure proper operation
That has no effect on the resistance. The current doubles also.
Ripples in electricity are usually defined as small, unwanted variations due to direct current. The effect of using a filter capacitor in this environment may vary, but usually has a smoothing effect on the ripple.