A 100 V battery is connected through a switch to two identical resistors and an

Question

A 100 V battery is connected through a switch to two identical resistors and an ideal inductor as shown in the figure. The resistor has a resistance of 100 Ohm. The inductor is a solenoid with n = 3990 turns/m, a cross-section area of 50 cm^2 and a length of 10 m. The switch is initially open. The switch is closed to 1. Immediately after the switch is closed to 1, what is the current in R? 5 ms after the switch is closed, what is the current in R? The switch is closed to 1 for a long time and disconnected from 1. Then the switch is closed to 2. Immediately after the switch is closed to 2, what is the current in R? 5 ms after the switch is closed to 2, what is the energy stored in the inductor L? The switch is closed to 1 for 10 ms and disconnected from 1. Then the switch is closed to 2. Immediately after the switch is closed to 2, what is the energy stored in the inductor L? 5 ms after the switch is closed to 2, what is the current in resistor R?

Explanation / Answer

inductance of solenoid
L = miuo*N^2*A/l
=(1.2566*10^-6)*(3990)^2 * (50*10^-4) / (10)
=0.01 H

a)
I = V/R * (1- e^(-R*t/L))
= 100/100 * (1- e^(-R*0/L))
=1 A
Answer: 1 A

b)
I = V/R * (1- e^(-R*t/L))
= 100/100 * (1- e^(-100*5*10^-3/0.01))
=1*(1- very small quantity)
Answer: 1 A

c)
I = V/R * e^(-R*t/L)
= 100/100 * e^(-R*0/L)
=1 A
Answer: 1 A

d)
E = 0.5*L*I^2
= 0.5*0.01*1^2
=0.005 J
Answer: 0.005 J

please ask only 4 parts at a time

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