Electromagnetism Problem 3 A battery with negligible internal resistance and wit

Question

Electromagnetism

Problem 3 A battery with negligible internal resistance and with electromotive force E-12.0 V is connected through a switch S to an inductor with inductance L 400 HH and two resistors with resistances RI 1.20 k and R 20.0 k2, see figure 3.1. The switch has been closed for a long time 3.1 Calculate the power dissipations PI and P2 in the two resistors and the stored energy UL in the inductor. At time t to the switch is opened, see figure 3.2 3.2 Calculate the voltage across R2 right after the switch is opened 3.3 Calculate how long time Mt goes until the current has decayed to half of the value at t t Figure 3.1 Figure 3.2

Explanation / Answer

This is an exercise of the circuit RL

Data

E = 12 V

L = 400 H = 400 10-6 H

R1 = 1.20 k= 1.20 103

R2 = 20.0 k = 20.0 103

Part 3.1)

current in the circuit is

I = Io (1- exp( -t/) )

= L/R1

Io= E/R1

when the switch is connected a long time, the exponential approaches zero

I = Io = E/R1 if R2 was infinite

The current in the resistor R2 is

E = I2 R2

I2 = E/R2

I2 = 12/ 20 103

I2 = 0.6 10-3 A

at the node

I = I1 +I2

I = E/R1

I = 12/1.2 103

I = 10 10-3 A

I1 = I- I2

I1 = 10 10-3 – 0.6 10-3

I1 = 9.4 10-3 A this is the current flowing through the branch L because of the effect of R2

We calculate the powers

P2 = E I2

P2 = 12 0.6 10-3

P2 = 7.2 10-3 W

P1 = V I1 = I12 R1

P1 = (10 10-3)2 1.2 103

P1 = 1.2 10-1 W

U = ½ L I2

U = ½ 400 10-6 9.4 10-3

U = 1.88 10-6 J

Part 3.2)

In the decay

I = I1 exp(-t/)

to open the switch t =0 s

I = I1

the voltage at R2

V2 = I1 R2

V2 = 9.4 10-3   20 103

V2 = 188 V

Part 3.3 )

The expression for the decay is

I = Io exp (-t/)

= L/R1

= 400 10-6 / 1.2 103

= 3.33 10-7 s

I/Io = exp ( -t/)

I = ½ Io

0.5 = exp (-t/)

ln 0.5 = -t/

t = - ln (0.5)

t = - 3.33 10-7 ln(0.5)

t = 2.31 10-7 s

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