(II) (a) In Fig. 30–28, assume that the switch has been in pos...

Question

(II) (a) In Fig. 30–28, assume that the switch has been in position A for sufficient time so that a steady current I₀ = V₀/R flows through the resistor R. At time t = 0, the switch is quickly switched to position B and the current decays through resistor R' (which is much greater than R) according to I = I₀ e⁻ᵗ/ʳ. Show that the maximum emf εₘₐₓ induced in the inductor during this time period is (R'/R)Vo. (b) If R' = 45R and Vo = 145 V, determine εₘₐₓ. [When a mechanical switch is opened, a high-resistance air gap is created, which is modeled as R' here. This Problem illustrates why high-voltage sparking can occur if a current-carrying inductor is suddenly cut off from its power source. The very high voltage can produce an electric field great enough to ionize atoms of air, which emit light when electrons recombine with the ions.]

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Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with LR circuits. This problem says in a circuit, a switch has remained in position P for a long time, establishing a state state current I knot is equal to V divided by R through resistor, R at T equal to zero. The switch is suddenly moved to position Q causing the current to decay through a much larger resistor. RL according to I is equal to I knot multiplied by E raised to the negative T divided by Tau determine the maximum emfe ma induced in the inductor during this transition. If RL is equal to 80 R and V is equal to 140 volts. Well, the question we're given in a circuit diagram of what was described in the problem. We're also given four possible choices. As our answers. For choice A we have 5.6 kilovolts. For choice B, we have 8.0 kilovolts. For choice C, we have 11 kilovolts and for choice D, we have 14 kilovolts. So we need to calculate our maximum induced EMF and to do that, we need to recall our formula for the induced EMF in an inductor. So if we call for that formula, we have E is equal to minus L multiplied by the derivative of I with respect to T, where E here is our induced emfl is our inductance I is the current and T is time. Now, we were given a formula for the current as a function of time. So we can actually plug that in and take a derivative. So we'll have E is equal to minus L multiplied by the derivative with respect to T of the quantity. And for I, we have, I not multiplied by E raised to the minus T divided by Tau. So we can take it RD of. And so we'll have E is equal to minus L, the I kno is a constant. So we will move through the derivative. So I'll have that multiplied the L multiplied by I knot. And then I'm gonna take the derivative of E raised to the minus T divided by Tau. And when I do that, I'll get minus one divided by Tau multiplied by E to the minus T divided by Tau. Now, I actually want the maximum induced DMF and that max actually occurs at T equal to zero because that exponential curve is gonna have its maximum value of one when T is equal to zero afterwards. Since it's uh raised to a negative exponent, that value is going to be less than one. So plugging in T equal to zero. Will get IMAX is equal to the negative signs cancel out. And so I'll just have L multiplied by I not divided by Tau. Now, in the problem, we were given a formula for I not in terms of V and R that we can plug in. And if you recall our time, constant Tau for an LR circuit, Tau is going to be equal to L divided by the resistance in the circuit which once we make this um switch flip over to point Q, that resistance is going to be RL. So Tau is gonna be equal to L divided by RL. So we can plug both of those expressions in and we'll have Emma is equal to L multiplying the quantity of V divided by R and that's gonna be divided by Tau, which is actually L divided by RL. We can simplify this bit. The LS will cancel and I will get E max is equal to. And here we'll move the one divided by RL up to the new meter. So we'll get V multiplied by RL, divided by R. So at this point, we can actually plug in values. So we'll have emacs is equal to for V that was 140 volts for RL, that was AD R and that quantity is gonna be divided by R. So the RS will cancel out and then un plug those values into my calculator. And doing so we'll get emacs is equal to 1.1 multiplied by 10 to the four volts. And here I've just rounded to two significant figures. However, if I look at my answer choices, they're all in kill volts. So I'll need to multiply this by the conversion factor of one kilovolt divided by 1000 volts since there's 1000 volts in 1 kg volt. So recalculate and we'll get Emma, it's equal to 11 kilovolts. So that's the answer that I'm looking for. And when I look at my answer choices, this corresponds to answer choice C So just to recap of what we've done here, we started off with our definition for the induced EMF or an inductor. And we plugged in the formula for the current that we're given since we had to take a derivative of the current to calculate our induced EMF. So once we took that derivative, we then set our time to equal to zero. Since that would give us the maximum induced EMF we then just had to simplify our expression and plug in values to calculate our final answer. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Induced EMF

Exponential Decay of Current

Voltage Division in Resistors

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