Is it possible to have a voltage drop across the energy storage element greater than the supply?

Is it possible to have a voltage drop across the energy storage element greater than the supply?

Yes, the voltage drop across the inductor or the capacitor in a series LCR circuit can be greater than the applied voltage of the AC source. This is because the voltage drops across these two elements are not phase voltages.

Can the voltage magnitude at resonance across any of the circuit components be greater than the magnitude of the source voltage?

-The voltage across the inductor can exceed the maximum source voltage. -The voltage across the capacitor can exceed the maximum source voltage. -None of these voltages can ever exceed the maximum source voltage. -Kirchoff’s loop rule is only valid for DC circuits, and does not apply to this AC situation.

Can voltage be higher than source?

Yes, even in passive circuits the output voltage can be greater than the input voltage. There are many examples, resonant circuits and transformers are two key example.

Can the peak voltage across the inductor be greater than?

Let a LCR circuit is connected across an AC supply with the emf E = E0 sin ωt. Therefore if magnitude of L R > 1 at resonance the value of the voltage across the inductor will bw greater than the peak value of the supply voltage.

Does voltage drop across inductor?

There will be a voltage across an inductor as the current in the inductor changes. Once the current reaches its steady-state value it will have zero voltage drop, because the current will not be changing.

Can the voltage drop across?

Voltage drop is the decrease of electrical potential along the path of a current flowing in an electrical circuit. Voltage drops in the internal resistance of the source, across conductors, across contacts, and across connectors are undesirable because some of the energy supplied is dissipated.

What is quality factor in LCR circuit?

Given below is the figure of an LCR circuit. The quality factor is the ratio of maximum energy stored in a capacitor or inductor to the energy dissipated by a resistor or we can say energy dissipated during one cycle of oscillation.

Why is input voltage is higher than output?

If the output voltage becomes higher than the input voltage, a large current may flow from the output to the input, which may damage the IC. This is caused by a parasitic diode in the internal MOSFET. The parasitic diode may turn on if the output voltage is higher than input voltage by its Vf.

Can voltage drop be higher than the voltage?

The simplest circuits, namely, circuits that just have a battery and a few resistors in the them, cannot have a voltage drop across any resistor that is higher than the battery voltage of a circuit.

What causes a voltage drop across an inductor?

The more rapidly current is decreased, the more voltage will be produced by the inductor, in its release of stored energy to try to keep current constant. When the switch is closed, the inductor will briefly oppose the change in current from zero to some magnitude, but will drop only a small amount of voltage.

What does an inductor do to voltage?

The effect of an inductor in a circuit is to oppose changes in current through it by developing a voltage across it proportional to the rate of change of the current. When there is a sinusoidal alternating current (AC) through an inductor, a sinusoidal voltage is induced.

How are voltage drops greater than source voltage?

Voltage drops are greater than source voltage. How is it possible? This circuit was given in my textbook. The source voltage is only 10 Volts, but the individual voltage drop across resistor and inductor are 6 Volts and 8 Volts respectively, which is greater than the source voltage.

How is the sum of your and L voltages equal?

Usual Kirchhoff’s laws holds also for phasors, but you need to use vector (or complex) addition instead of conventional real number addition. R and L are in series, so the vector sum of the voltages across them equals the overall voltage across the series (10V 50Hz).

How to calculate the phase of a voltage drop?

Using the vector addition you find out that the voltage across the series has an amplitude which is the hypotenuse of a triangle having 6V and 8V as sides, hence: Of course this is only the amplitude of the voltage. You can use trigonometry formulas to get the phase.

How can a voltage across a capacitor be greater than the voltage?

Eventually the backwards voltage across the inductor causes the inductor current to go to zero. At that point, all the energy originally in the inductor when the switch closed has been transferred to the capacitor. This process can be repeated to reach arbitrarily large voltages on the capacitor in theory.