Question

The switch in the figure below is open for t<0 and is then thrown closed at time t = 0. Assume R = 5.00 0, L = 4.00 H, and 8 - 16.0 V. Find the following as functions of time thereafter. (Do not enter units in your answers. Assume current is in A and time is in s Use the following as necessary: t.) R 2R w R ele (a) the current in the inductor 13 (b) the current in the switch

Answer 1

At t = infinity, the resistance of L is 0. The net resistance if L is short circuited is

R+1/(1/2R+1/R)=5R/3

The total current throught the battery at t = infinity is therefore

E/(5R/3) = 3E/5R

Current through 2R resistance is at t = infinity

(3R/5R) R/(3R) = E/5R

Therefore the current through inductor at t = infinity is E/5R

The resistance as seen by the inductor is

2R+1/(1/R+1/R) = 5R/2

Therefore the time constant is 5R/2L

The current through inductor is therefore

$I_L = \frac{E}{5R}(1-e^{-t/(2L/(5R)))})$

substitute

E = 16 V

R = 5 ohms

L = 4 H

$I_L = \frac{16}{25} \left(1-e^{-\frac{25 t}{8}}\right)$

or

$I_L = 0.64\left(1-e^{-3.125 \,t}\right)$

The unit is Amperes

b) The current flowing through R in parallel with switch is

Thus the current through the switch is

$I_s=E/2R-E/10R (1-e^{-5Rt/2L})+I_L=+\frac{E}{2 R}+\frac{E \left(1-e^{-\frac{5 R t}{2 L}}\right)}{10 R}$

substitute the values

E = 16 V

R = 5 ohms

L = 4 H

$I_s=\frac{8}{5}+\frac{8}{25} \left(1-e^{-\frac{25 t}{8}}\right)$

or