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behavior of LR circuit An LR circuit for Problem 1. Close Si while leaving S2 op

ID: 2255705 • Letter: B

Question

behavior of LR circuit

An LR circuit for Problem 1. Close Si while leaving S2 open: Through what elements does the current flow (battery? resis­tor? inductor?)? In circuits with just resistors, we tacitly assumed that the current went from zero to some value the instant a switch was closed. This is impossible in this case. Why? Since it cannot jump immediately, the current at the instant the switch is closed must be zero. What is the potential drop across the resistor (Va - Vb) and across the inductor (Vb - Vc)? At the instant the Si is closed, the current is zero. Is it possible for dt/dt # 0? If so, what is it. If not, why not? We know from experience (your lab!) that eventually the current will flow. Thus, some small amount of time after the switch is closed, the current must be some value i(t = 0 4- dt) which we will call i 1. What is the potential drop across the resistor (Va - Vb) and across the inductor (Vb - Vc) in terms of il? What is dt/dt t=0+dt in terms f oil? What is the sign of d2i/dt2 at time t = 0+cdt? To get this compare the values To see this question, go to d2L under quizzes and enter your results for the last few parts. If your answers are correct, you will be able to see a pdf with this part of the question under "Content rightarrow Homework" on d2L. Discuss the magnetic field of the inductor and the energy therein as a function of time. Is the stored energy increas­ing or decreasing? What is the direction of ?

Explanation / Answer

a
i. battery,Resistor,inductor all three


ii.impossible in this case...as current through inductor can't change instantaneously bcz

across inductor as e = Ldi/dt so instantaneous change of current means infinite voltage(as instantaneous change means dt->0),which is not possible.


iii. across resistor = iR = 0*R = 0
now,by kirchoff law-

e = iR + voltage across inductor

=> e =0 + voltage across inductor

=> voltage across inductor = e


iv. as e = Ldi/dt

as e is not equal to zero

so di/dt also not equal to zero


v. across resistor = i1R

now,by kirchoff law-

e = iR + voltage across inductor

=> e =i1R + voltage across inductor

=> voltage across inductor = e