A 4.00-mF capacitor is connected in series with a 7.00-mH inductor. The peak current in the wire

A 4.00-mF capacitor is connected in series with a 7.00-mH...

A $4.00-\mathrm{mF}$ capacitor is connected in series with a $7.00-\mathrm{mH}$ inductor. The peak current in the wires between the capacitor and the inductor is $3.00 \mathrm{~A}$. a) What is the total electric energy in this circuit? b) Write an expression for the charge on the capacitor as a function of time, assuming the capacitor is fully charged at $t=0 \mathrm{~s}$.

A $4.00-\mathrm{mF}$ capacitor is connected in series with a $7.00-\mathrm{mH}$ inductor. The peak current in the wires between the capacitor and the inductor is $3.00 \mathrm{~A}$.
a) What is the total electric energy in this circuit?
b) Write an expression for the charge on the capacitor as a function of time, assuming the capacitor is fully charged at $t=0 \mathrm{~s}$.

Key Concepts

LC Circuit Oscillations

An LC circuit consists of a capacitor and an inductor connected together, and it exhibits oscillatory behavior due to the continuous energy exchange between the electric field of the capacitor and the magnetic field of the inductor. The natural frequency of these oscillations is determined by the values of the capacitor and inductor, and it is given by ? = 1/?(LC), a key concept when analyzing the dynamics of these circuits.

Capacitor Energy Storage

A capacitor stores energy in the electric field created between its plates when it is charged. The energy stored in a capacitor is given by the expression U = (1/2)Q^2/C or equivalently U = (1/2)CV^2. This concept is crucial for understanding how electric energy is accumulated in the circuit and how it is later converted into magnetic energy in the inductor.

Inductor Energy Storage

An inductor stores energy in its magnetic field when current flows through it. The energy stored in an inductor is given by U = (1/2)L I^2. This principle is central to analyzing the system’s behavior, as the inductor’s ability to store and release energy is what sustains the oscillations in the LC circuit.

Simple Harmonic Motion in LC Circuits

The oscillatory behavior of an LC circuit can be described by simple harmonic motion, where the charge on the capacitor and the current in the inductor vary sinusoidally with time. The differential equations governing the system yield solutions that are typically sinusoidal functions, such as cosine or sine functions, which describe how the charge on the capacitor changes with time under specific initial conditions.

Transcript

00:01 Capacitance c equals to 4 .00 milliferaid that is equals to 4 .00 multiplied by 10 to the power minus 3 ferret and inductor l equals to 7 .00 millie henry that is 7 .00 multiplied by 10 to the power minus 3 henry.

00:18 Okay and the peak current in the wires between capacitor and inductor is i max equals to 3 .0 amp.

00:27 Okay so now for the part a of the question we have have to determine the total electrical energy in this circuit.

00:33 So the total electrical energy, this will be ue and it will be given by 1 by 2 cv square or it can be written as 1 by 2 i -l i max square.

00:45 So substituting values in this equation, so we will get ue equals to 1 by 2 multiplied by l which is 7 .0000 multiplied by 10 to the power minus 3, multiplied by i max which is 3 .00 ampere, so whole square.

01:00 So from here after solving electrical energy ue comes out to be 0 .0315 jule or it can be written as 31 .5 millijoule.

01:11 So this is the expression for this is the answer for the electrical energy in the circuit.

01:16 Now moving to the part b in which we have to determine the expression for the charge on the capacitor as a function of time.

01:25 So, the expression will be given by q equals to q0 co0 coox, omega not modpled by t, okay? where omega not is the frequency for the lc oscillator and it is given by omega not equals to 1 by under root lc and q0 is the maximum charge and from the energy conservation we can write that 1 by 2 q0 squared by c, it will be equals to 1 by 2 l .i max square.

01:53 So from here after rearranging, we get q0 equals to imex multiplied by under root lc.

02:01 So substituting values in these two expressions, 1 and 2, so to get omega not value and q not value...

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