3. In the circuit shown below e-9v, R1-15 Ohms, and L-0.1H. Switch S is closed a

Question

3. In the circuit shown below e-9v, R1-15 Ohms, and L-0.1H. Switch S is closed at t-o s. Just after the switch is closed: Part I a) What is the potential difference (Voc-Va-V) across the inductor? b) Potentials at points a and b are in the following relationships i) Point a is at higher potential than b Point b is at higher potential than da The potentials are equal; Not enough information. ii) iii) iv) Part II. c) What is the maximum value of current through Rs at sufficiently longer times (i.e.very lon d) e) time after the switch was closed)? How much time (t60%) is needed to reach 60% of that (maximum) value? How much energy is stored within the inductor at that moment of time? V.

Explanation / Answer

3. (A) Just after switch is closed,

I = 0

so VR = 0

KVL: e - VR - VL = 0

VL = Vb -Vc = e = 9 Volt .........Ans

(B) Ans(ii) Point B is at higher potential.

(C) after long time,

I0 = e / R1 = 9 / 15 = 0.6 A

(D) I = I0 [ 1 - e^(-t/T)]

0.60 I0 = I0[ 1 - e^(-t/T)]

t = - T ln(0.40)

T = L/R

t = - 0.1 ln(0.40) / 15 = 6.11 x 10^-3 s Or 6.11 ms

(E) I = 0.60 x 0.6 = 0.360 A  

and energy stored = L I^2 / 2

= (0.1 H) (0.360^2)/2

= 6.48 x 10^-3 J

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