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Capacitative currentLeak current (out)1 1 1 J2040 60 Time (us)80INaEarly currentIKout Late currentTime (ms)92001 Sinauer Associates...

Question

Capacitative currentLeak current (out)1 1 1 J2040 60 Time (us)80INaEarly currentIKout Late currentTime (ms)92001 Sinauer Associates

Capacitative current Leak current (out) 1 1 1 J 20 40 60 Time (us) 80 INa Early current IKout Late current Time (ms) 92001 Sinauer Associates



Answers

The current $i$ in Fig. $14.23$ is (A) $\frac{1}{5} \mathrm{~A}$ (B) $\frac{1}{10} \mathrm{~A}$ (C) $\frac{1}{15} \mathrm{~A}$ (D) $\frac{1}{45} \mathrm{~A}$

Hello, friends. Here it is given in the given circuit attitudes culture of Europe Current is I not? We have toe find that time eight at which current becomes I not by two toe equivalent resistance of the circuit will be R one plus Otto, that is 200 plus 300 500. 0, so it is the question based on Dickie off current through the inducting quite so decay off current through the inducting while is given by ice called. So I not explanation off are equivalently upon area. So this value is given I not by two I not explanation off Our equivalent are equivalent is 500 and took t upon el. That is 50,000,100 on solving it value off the we will get 6.5 Sorry. Six point night and to tend to the bar minus five seconds. That's all. Thanks for watching it

Right. So in this problem, if we close the switch ex ally, so want to know the current inside the this branch. So because our one are two, they're connected ing paranoid. All right, so the body across our one equal to this, that 1 20 volts. So the currents are one I want equal VD belie are one. Okay, so there's a 1 20 volts Dubai by 1 20 homes, which is why I'm here, okay?

The current at the time he is given by I. T. Is equal to P by uh multiplication one minus E. To the power minus Pr by capital. And the study state value I know difficult to PVR and I can also write the value I buy I note is equal to N. Is equal to one minus E. To the power minus we are by capitol L solving it further, I can idea expression of U. To the power minus T. R. By L is equal to one minus M. So be not. R. Y. L is equal to learn one by one minus. And finally I can write the value of peanut is equal to L by our learn one by one minus end. Which one solving I get the answer at one point 49 seconds.

On the topic off self induct INTs and in doctors were given a graph off the current flowing through an in doctor with time. Now we're given the resistance off Thean doctor, and we asked to calculate the voltage across the in doctor at different times. So before we calculate the voltage, we need an expression on the voltage across and in. Doctor, we know is minus al times. The i d t. Where l is the induct INTs off that conductor and the i d. T is the rate of change of current. So what we will be doing in this question is taking the ingredient off. The straits slopes the radiant off the straight parts. What's the curves to calculate the i d. T. So essentially working out Delta I by Dr T. So for part A were asked to calculate this voltage. A T is equal to milliseconds, so the T equal to too many seconds v out of using the equation above his minus and the induct. Ince's constant at five million Henry's or five times 10 to the minus three mhm multiplied by the change in current any change in time, which is six amperes minus to em peers divided by the change in time three minus one. And that's milliseconds. So that's times 10 to the minus three. And this gives us the voltage across the in dr two B minus 10 vaults. Now let's do the same. AT T is equal to formally seconds. So now we want to calculate the potential difference across the conductor at a different time. So we all again is minus l, which is constant. Five. I'm staying to the minus three Henry's times. D i D t. Which for this portion of the graph is foreign piers minus exam piers divided by Dr T five minus T times 10 to the minus three seconds. Which gives us the voltage across the in doctor at 40 seconds to be five votes. And finally we'll do the same. Yeah, And finally, for part c, we have Time t is equal to eight milliseconds and the voltage across the in doctor at this time is again minus l, which is minus five times 10 to the minus three. Hendry's comes the I d t. Which is the change in current one minus three and peers over the change in time off 10 minus six milliseconds. So that's time to 10 to the minus three, and hence we get the voltage across the in doctor at eight milliseconds to be 2.5 Vault's.


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