A coil with an inductance of 2.0 H and a resistance of 10Ωis suddenly connected to an ideal batt

A coil with an inductance of 2.0 H and a resistance of 10Ωis...

A coil with an inductance of 2.0 $\mathrm{H}$ and a resistance of 10$\Omega$ is suddenly connected to an ideal battery with $\mathscr{E}=100 \mathrm{V}$ . At 0.10 $\mathrm{s}$ after the connection is made, what is the rate at which (a) energy is being stored in the magnetic field, (b) thermal energy is appearing in the resistance, and (c) energy is being delivered by the battery?

A coil with an inductance of 2.0 $\mathrm{H}$ and a resistance of 10$\Omega$ is
suddenly connected to an ideal battery with $\mathscr{E}=100 \mathrm{V}$ . At 0.10 $\mathrm{s}$
after the connection is made, what is the rate at which (a) energy is
being stored in the magnetic field, (b) thermal energy is appearing
in the resistance, and (c) energy is being delivered by the battery?

Key Concepts

Electrical Power Delivered by a Battery

The battery supplies electrical energy to the circuit at a rate determined by the product of its electromotive force and the instantaneous current (P = ? i). This concept links the energy provided by the source to the energy stored in components and the energy dissipated as heat, forming the basis for energy conservation and power management in electrical circuits.

Thermal Energy Dissipation in a Resistor (Joule Heating)

Resistors dissipate electrical energy in the form of heat according to the power law P = i^2 R. This conversion of electrical energy to thermal energy is described by Joule's law and is a key aspect of energy balance in circuits. Understanding this concept is important for evaluating energy losses and efficiency in electrical systems.

Transient Analysis of RL Circuits

This concept involves understanding the time-dependent behavior of current in an RL circuit, which is governed by a first-order differential equation. The solution to this equation shows that the current grows exponentially toward its steady-state value with a time constant given by the inductance divided by the resistance (? = L/R). This analysis is crucial for understanding how rapidly the system reaches its steady state after a sudden change, such as the closing of a switch.

Energy Storage in an Inductor

Inductors store energy in their magnetic fields, and the stored energy is given by the formula (1/2) L i^2. The rate at which this energy changes, determined by taking the time derivative, is essential for analyzing how quickly energy is being stored or released in dynamic circuits. This concept is fundamental in understanding transient energy relationships in circuits containing inductors.

Transcript

00:01 So here for part a, from equation 3049 and equation 3041, we can then find the rate at which the energy is being stored in the inductor.

00:14 So we can say the change in potential energy of the inductor with respect to time would be equal to the derivative of one half l .i squared.

00:26 This would be, again, with respect to t.

00:28 And this would be the inductance times the current times the change in current with respect to time.

00:35 This would be equal to the inductance multiplied by the emf divided by the resistance r.

00:42 This is from ums law.

00:43 And then the change in the current with respect to time, this would simply be 1 minus e to the negative to the negative time power divided by the inductive time constant, tau sub l.

01:01 At this point, we can then multiply this by the emf divided by r.

01:08 This would be multiplied by 1 over the inductive time constant.

01:13 So we're simply taking the derivative and then multiplied by e to the negative t over tau subl.

01:22 Close parentheses, this would be multiplied.

01:25 And we can then say that for part a, the change, essentially the rate at which the inductive, energy is being stored in the inductor with respect to time would be equal to the emf squared divided by the resistance r multiplied by 1 minus e to the negative t over tau sub l again the inductive time constant multiplied by e to the negative t divided by tau sub l and we know that the inductive time constant is equaling the inductance divided by the resistance this is equaling 2 .0 henries divided by 10 ums, this is equaling 0 .20 seconds.

02:12 Therefore, we also know that the emf is equaling 100 volts.

02:17 So we can then say that when t is equaling 0 .10 seconds, the inductive time constant, rather the change at which the rate is being stored in the inductor with respect to time would be equal to 2 .4 times 10.

02:36 To the second watts and again this would be the equation that you're plugging these these variables rather into so this would be essentially our answer for part a for part b now we want to use equation 2622 and equation 30 41 the rate at which the resistor is generating thermal energy we can say p thermal would be equal to the current squared times the resistance.

03:14 This would be equal to the emf squared divided by the resistance squared multiplied by 1 minus e to the negative t over tau sub l...

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