I have a few questions i missed and scored a 69 on my lesson. this is one point below passing grade of 70. i need help with the questions i missed. i have been rereading this lesson over and over but cannot seem to locate the correct answers. i am going to post the questions here, i hope someone can help me out.
Question 26 (which i did get right) is
*A step voltage of 15 volts is applied to a 100 ohm resistor in series with a 2-H inductor. The initial current rise (near time 0) equals.
(1) 0.133 A/s
(2) 0.15 A/s
(3) 6.67 A/s
(4) 7.50 A/s (my answer)
(5) none of these
Questions 27 to 32 refer to the circuit of question 26 (the following questions are the ones i am having issue with)
Question 28) the steady state value of voltage across the inductor is
(1) 0 V
(2) 0.15 V
(3) 6.67 V
(4) 7.5 V
(5) 15 V (my answer)
Question 29) the intial value of voltage across the inductor is
(1) 0 V (my answer)
(2) 0.150 V
(3) 6.67 V
(4) 7.50 V
(5) 15 V
Question 31) After one time constant the voltage across the inductor is.
(1) 5.5 V
(2) 6.3 V
(3) 4.5 V (my answer)
(4) 9.5 V
(5) 15 V
question 32) after one time constant the current through the inductor is.
(1) 37 mA
(2) 63 mA
(3) 55 mA
(4) 95 mA
(5) 150 mA (my answer)
Question 33) A 1000 V step is applied across a 400 k ohm resistor in series with a 2uf capacitor. the initial rate of rise of voltage across the capacitor is found by V divided by T, where T= R x C. the inirtial rate of rise is.
(1) 0.25 v/s
(2) 500 V/s
(3) 1000 V/s
(4) 1250 V/s (my correct answer)
question 34) the steady state value of voltage across the resistor, question 33, is
(1) 0 V
(2) 500 V
(3) 1000 V (my Answer)
(4) 1250 V
any help would be greatly appreciated. i will complete the course if i can just correct and resubmit this exam with a 70 or better grade.
thanks
Robert,
Since you can submit the corrected quiz for a grade I will not provide the answers but will try to help you reason through this.
A step voltage applied to the circuit would be seen as an instantaneous change in voltage by the inductor. The inductor works to oppose CURRENT change and will not initially pass current. It is a bit tricky since they use the phrase "near time 0" which leaves open several solutions unless this is taken as instantaneously after the step change in which case you can solve for di/dt from v = L(di/dt) which you apparently did.
28. After the transient dies out the circuit is in steady state and the inductor no longer acts to oppose the change in current since the current is now constant. The voltage across the inductor given by v = L(di/dt) can be calculated using di/dt for no current change.
29. The inductor opposes a CHANGE in current and does so by allowing the voltage across the inductor to change to keep the current constant. In order to do so the voltage across the inductor will have to adjust so the current in the circuit remains zero. Figure what the voltage across the inductor needs to be to keep the current zero.
31. One time constant results in a CHANGE of 63% from the initial to the final value which in this case is 63% of the initial voltage minus the final voltage subtracted from the initial voltage. Think this one through carefully especially with what is the initial value and what is the final value and what is the change from where to where.
32. From 31 you know the voltage across the inductor and using circuit analysis you can calculate the voltage across the resistor and then get the current.
34. When in steady state the current in the circuit will be zero so you can calculate the voltage across the R.
Try to keep in mind some key points...The voltage across an inductor changes to oppose a change in current, the current through a capacitor changes to oppose a change in the capacitor voltage, and one time constant is 63% of the change from initial to final value.
Hope this helps.
Neil