Question

The circuit in the diagram consists of a 17.7 22 resistor, an inductor, and a 23.3 V battery. The inductor is made from a solenoid with the following parameters: 4V. 0000 Cross-Section Area = 16.9 cm2 Length = 3.6 cm Number of turns of wire = 1255 Part A Calculate the inductance of the inductor. Part B Calculate the time constant for the circuit. Part How long will it take for the current from the battery to reach 85% of its final value after the switch is closed? t = Part D When the current has reached 85% of its final value, what will be the back emf of the inductor at that time? Enter a positive value EBACK Part E What will be the maximum stored energy in the inductor?

Answer 1

3.6 X 10-2 al self Inductance, L= MONZA – 40X16*x (1255) x 16.98109 on, L= 0.093 H — Ans) b) Time constant, le=48= 0.0003 = 5.25 x 1023 second-fire) e) As the switch is closed, L de tir = V -O solving this equation we get, i(t) = 27-1-e-&t}— where, 1o -Vk when liſty is 85% of final value , lic(t) = to -5.25X1 opy e $352169 – 1-. - o = 20 or, t= -5.25X10-3x In[Yao) a t= 9.96X10-3 second-thong , d) emf of inductor, Ve=l di = RI.e-&t=vesen when, i = 857. I, , then, t= 9.96X1003 Seconda 50, VI = 23:3x @ 5:25 x 10-3 op, V.= 349 Volt - Ame) This is Enaey- es Manimum stored energy of Inductor, Dons ŽLICOM op, Umane § - To= 3X0.093 (23:3)2 [:'To=987 op, T Umană 0.08 Joule — Ame) man mu