
Which energy is stored in the inductor and capacitor?
Answer
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Hint:A capacitor is an electrical component with two terminals that can store energy in the form of an electric charge. It's made up of two electrical wires that are separated by a specified amount of space. Inductors are widely used to lessen or control electric spikes by temporarily holding energy in an electromagnetic field and then releasing it back into the circuit. Now we'll look at how much energy is stored in an inductor and a capacitor.
Complete step by step solution:
In a capacitor, the energy is stored in the form of electrostatic energy. In an inductor, the energy is stored in the form of magnetic flux.
Energy stored in capacitor:
Electrical potential energy is stored in a capacitor and is thus related to the charge \[Q\] and voltage \[V\] on the capacitor. When using the equation for electrical potential energy \[\Delta qV\] to a capacitor, we must be cautious. Remember that a charge q passing through a voltage \[\Delta V\] has a potential energy of \[\Delta PE\]. The capacitor, on the other hand, begins with no voltage and progressively increases to its full value as it is charged. Because a capacitor has zero voltage when it is uncharged, the first charge placed on it causes a voltage change of \[\Delta V = 0\]. Because the capacitor now has its full voltage V, the final charge is placed on its experiences \[\Delta V = V\].
During the charging process, the average voltage on the capacitor is $\dfrac{V}{2}$.
So, the average voltage experienced by the full charge \[q\] is $\dfrac{V}{2}$.
Thus, the energy stored in a capacitor, \[{E_{cap}}\]
${E_{cap}} = \dfrac{{QV}}{2}$
where \[Q\] represents the charge on a capacitor when a voltage V is applied. (It's worth noting that the energy is $\dfrac{{QV}}{2}$, not \[QV\]).
Because the capacitance \[C\] of a capacitor is connected to charge and voltage by \[Q = CV\], the expression for \[{E_{cap}}\] can be algebraically manipulated into three equivalent expressions:
${E_{cap}} = \dfrac{{QV}}{2} = \dfrac{{C{V^2}}}{2}\dfrac{{{Q^2}}}{{2C}}$
Where\[Q\] is the charge on a capacitor \[C\] and \[V\] is the voltage. For a charge in coulombs, the voltage in volts, and capacitance in farads, the energy is measured in joules.
Energy stored in Inductors:
The magnetic field in an inductor stores energy when an electric current flows through it. In the case of a pure inductor \[L\], the instantaneous power required to start the current in the inductor is,
$P = iv = Li\dfrac{{di}}{{dt}}$
As a result, the integral gives the energy input to build up to a final current \[i\].
Energy stored $ = \int\limits_0^t {Pdt} = \int\limits_0^I {Li'di' = \dfrac{1}{2}L{I^2}} $
Note: We know that we need to employ a diode to block the flow of current in a specified direction.
We utilize a capacitor to block DC if we need to.
An inductor is used to block extremely high-frequency AC.
We can utilize resistors, capacitors, and inductors to create a filter (and op-amps and transistors, etc.)
Capacitors, inductors, and diodes are used to create a switch-mode power supply.
A large capacitor is used to reduce ripple voltage on a power supply. We could also use an inductor if we need to reduce ripple even further.
Complete step by step solution:
In a capacitor, the energy is stored in the form of electrostatic energy. In an inductor, the energy is stored in the form of magnetic flux.
Energy stored in capacitor:
Electrical potential energy is stored in a capacitor and is thus related to the charge \[Q\] and voltage \[V\] on the capacitor. When using the equation for electrical potential energy \[\Delta qV\] to a capacitor, we must be cautious. Remember that a charge q passing through a voltage \[\Delta V\] has a potential energy of \[\Delta PE\]. The capacitor, on the other hand, begins with no voltage and progressively increases to its full value as it is charged. Because a capacitor has zero voltage when it is uncharged, the first charge placed on it causes a voltage change of \[\Delta V = 0\]. Because the capacitor now has its full voltage V, the final charge is placed on its experiences \[\Delta V = V\].
During the charging process, the average voltage on the capacitor is $\dfrac{V}{2}$.
So, the average voltage experienced by the full charge \[q\] is $\dfrac{V}{2}$.
Thus, the energy stored in a capacitor, \[{E_{cap}}\]
${E_{cap}} = \dfrac{{QV}}{2}$
where \[Q\] represents the charge on a capacitor when a voltage V is applied. (It's worth noting that the energy is $\dfrac{{QV}}{2}$, not \[QV\]).
Because the capacitance \[C\] of a capacitor is connected to charge and voltage by \[Q = CV\], the expression for \[{E_{cap}}\] can be algebraically manipulated into three equivalent expressions:
${E_{cap}} = \dfrac{{QV}}{2} = \dfrac{{C{V^2}}}{2}\dfrac{{{Q^2}}}{{2C}}$
Where\[Q\] is the charge on a capacitor \[C\] and \[V\] is the voltage. For a charge in coulombs, the voltage in volts, and capacitance in farads, the energy is measured in joules.
Energy stored in Inductors:
The magnetic field in an inductor stores energy when an electric current flows through it. In the case of a pure inductor \[L\], the instantaneous power required to start the current in the inductor is,
$P = iv = Li\dfrac{{di}}{{dt}}$
As a result, the integral gives the energy input to build up to a final current \[i\].
Energy stored $ = \int\limits_0^t {Pdt} = \int\limits_0^I {Li'di' = \dfrac{1}{2}L{I^2}} $
Note: We know that we need to employ a diode to block the flow of current in a specified direction.
We utilize a capacitor to block DC if we need to.
An inductor is used to block extremely high-frequency AC.
We can utilize resistors, capacitors, and inductors to create a filter (and op-amps and transistors, etc.)
Capacitors, inductors, and diodes are used to create a switch-mode power supply.
A large capacitor is used to reduce ripple voltage on a power supply. We could also use an inductor if we need to reduce ripple even further.
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