Any circuit using a capacitor will not work if the cap is short-circuited.
Does a magnetic field have an effect on a capacitor when it is placed between the plates? Yes, a magnetic field between the plates of a capacitor would have some effect. Without more information it is difficult to determine how much.
A capacitor try to leads the current while a inductor tries to legs the current so they cancels each other's effect ....
In order for a capacitor to pass current, the voltage across it must be changing. In a DC circuit, the voltage does not change so, at equilibrium, the capacitor is effectively an open circuit. We also call this DC blocking. You can take a signal with DC bias on it, perhaps because it came from a class A BJT amplifier, couple it with a capacitor, and the signal will make it through, but the DC bias will not.
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.
it will improve the power factor... The angle between voltage and current will decrease depends on capacitor value.
An open circuit, by definition, has no continuity, therefore there is no current flow. A failed capacitor in an open circuit would have absolutely no effect.
The force on capacitor plates in an electric circuit causes them to store electrical energy by creating an electric field between the plates. This results in the accumulation of electric charge on the plates, which can be released to power devices in the circuit.
The effect of a charged capacitor on the resistance in a circuit is that it can lead to a temporary increase in current due to the initial discharge when connected to a resistor, while an uncharged capacitor behaves as an open circuit at the moment of connection. Over time, as the charged capacitor discharges, the current decreases exponentially until it reaches zero, effectively behaving like a resistor with a time-dependent resistance. In contrast, an uncharged capacitor will not allow current to flow until it starts charging, resulting in a different initial resistance characteristic. Overall, the capacitor's state (charged or uncharged) influences how it interacts with the resistance in the circuit.
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 the experiment of flashing and quenching of a capacitor, the neon bulb twinkles because the charging and discharging of the capacitor cause the voltage across the capacitor to fluctuate rapidly. These fluctuations can cause the neon bulb to turn on and off, leading to the twinkling effect.
The magnetic field between capacitor plates does not have a significant effect on the overall performance of the capacitor. The main factors that affect a capacitor's performance are its capacitance, voltage rating, and dielectric material.
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.
The differential equation for a capacitor is dv/dt = i/c. Set that up in a circuit and force an AC power source, such as sin(theta), and you will see that lowering the frequency will increase the equivalent resistance. I'll leave that exercise for you. The net result is that a series capacitor is a high-pass filter, while a parallel capacitor is a low-pass filter.
You use a capacitor to store electrostatic energy. You use an inductor to store electromagnetic energy. You use a resistor to dissipate electrical energy.
It must (i) increase, or (ii) decrease, or (iii) stay the same. If (iii), there's no point in having it at all, so consider what happens to stage gain at the lowest frequencies, as the capacitor has less and less effect on the circuit.
Capacitors in parallel simply add up, similar to resistors in series... CTOTAL = sumI=1-N (CI) Capacitors in series work like resistors in parallel... CTOTAL = 1 / sumI=1-N (1 / CI)
Does a magnetic field have an effect on a capacitor when it is placed between the plates? Yes, a magnetic field between the plates of a capacitor would have some effect. Without more information it is difficult to determine how much.