A 1.50 mH inductor is connected in series with a dc battery of negligible internal resistance, a

step 1 This problem, we have an RL circuit with a switch. We're given values of 0 .750 kilooms for the

A 1.50 mH inductor is connected in series with a dc battery...

Key Concepts

Solving Exponential Equations

Many transient analysis problems in RL circuits require solving equations involving exponential functions. Techniques such as taking the natural logarithm of both sides of an equation are essential for isolating the time variable. This method allows one to solve for the time required to reach a certain fraction of the maximum current or energy, which is a common task in transient circuit analysis.

Exponential Growth in RL Circuits

The current in an RL circuit grows according to an exponential law given by i(t) = I_max (1 - e^(-t/?)), where I_max is the maximum current. This exponential behavior is a direct consequence of the first-order differential equation governing the circuit. The exponential function defines how quickly the circuit responds to changes, which is crucial when calculating specific timings like reaching half of the maximum current.

Energy Storage in Inductors

An inductor stores energy in its magnetic field when current flows through it, and the energy is given by the formula U = (1/2) L i^2. As the current increases exponentially in an RL circuit, the energy stored in the inductor also increases following the square of the current. Understanding this energy storage is important for analyzing the transient behavior and energy dynamics in circuits containing inductive components.

RL Circuit

An RL circuit consists of a resistor (R) and an inductor (L) connected in series. When a voltage source is applied, the current in the circuit does not instantly reach its maximum value but increases gradually due to the inductor’s opposition to changes in current. This is a typical example of first-order transient behavior in electrical circuits.

Time Constant (?) in RL Circuits

The time constant ? is defined as the ratio L/R, where L is the inductance and R is the resistance. It represents the characteristic time over which the current in the circuit approaches its steady state value. At time t = ?, the current reaches approximately 63.2% of its final steady state value, and it is a key parameter in determining the speed of the transient response.

Transcript

00:03 This problem, we have an rl circuit with a switch.

00:06 We're given values of 0 .750 kilooms for the resistance and 1 .50 milly henries for the inductance.

00:15 First part is asking what time the current reaches its half maximum value.

00:21 So to do this, we're going to need to recall that for an rl circuit, the current is a function of time.

00:29 It's given by v over r times 1 minus e to the minus r over l times t.

00:40 And recall that this is just an exponential behavior.

00:45 It starts out at zero and it levels off to our value of v over r.

00:51 So to find our, the time that it reaches half maximum value, we want to know first our maximum is given by this, current when t goes to infinity, which is just v over r.

01:13 And we want to solve the equation i, some t star that we don't know, we want to solve for that, equals i max over 2.

01:29 Well, this, if we plug in what both the left -hand side and right -hand sides are, this is going to be v over r, 1 minus e to the minus r over l times our little t star.

01:46 And then this equals v over 2r, our imacs over 2.

01:52 So we see that our v over r is cancel.

01:57 This equation is now going to be 1 minus e to the minus r over l times our t star equals 1 half.

02:10 This is going to be e to the minus r over l t star equals one half.

02:21 And here we want to take the natural log of both of these sides.

02:26 So we'll have, let me see where should i do this? i guess i'll just do that here.

02:35 So natural log of e to...

02:42 Remember that these two functions are inverses.

02:45 So this is just going to be minus r over l times t star.

02:52 And we know that natural log of one half is ln of 1 minus ln of 2.

03:00 And natural log of 1 is 0.

03:06 So we have minus r over l times t star equals minus natural log of 2.

03:17 So solving this, we have t.

03:20 Star our half maximum time is going to be l over r times ln of two and now let's see what this is so if we plug that in we have a 1 .5 milliom inductor 10 to the negative 3 and a 0 .750 resistance kiloom resistance so you want to put both of these into to omes and henries.

03:58 And multiplied by allen of two.

04:01 And plugging this in, we get 1 .39 times 10 to the negative 6 seconds.

04:15 For the second part, we want to find the time it takes for the stored energy to reach half its maximum value.

04:26 So i'm going to move this over here.

04:35 Just so you can still see it...

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