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Example 5.1:

The switch in the above circuit has been closed for a long time before it is opened at t=0. Find:

a) iL(t) for t>0-;

b) i0(t) for t>0-

c) v0(t) for t>0-

d) The percentage of the total energy stored in the 2 H inductor that is dissipated in the 10 resistor.

Solution:

a) The switch has been closed for a long time prior to t=0, so the voltage across the inductor must be zero at t=0-. Therefore the initial current in the inductor is 20 A at t=0-. Hence, iL(0

+) also is 20 A because an instantaneous change in the current cannot occur in an inductor. Replacing the resistive circuit connected to the terminals of the inductor with a single resistor of 10 : =+= 10)10||40(2eqR

The time constant of the circuit is L/Req, or 0.2 s, giving the expression for the inductor current as

Aeti tL520)( = , 0t .

b) The current in the 40 resistor is calculated by using current division; that is:

4010

100 +

= Lii

Note that this expression is valid for t>0- only (why). Then: Aeti t50 4)(

= , + 0t . c) By direct application of s law, the voltage v0:

Veitv t500 16040)(== , + 0t .

d) The power dissipated in the 10 resistor is:

Wev

tp to 102

256010

)( == , + 0t

The total energy dissipated in the 10 resistor is

Jdtetw t 2562560)(0

10 ==

.

t

)(

The initial energy stored in the 2 H inductor is

JLiw 400)0(2

1)0( 2 == .

Therefore the percentage of energy dissipated in the 10 resistor is

Percentage dissipated = 256/400 (100)= 64%.

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Example 5.2:

The switch has been in position (a) for a long time. The switch then moves to the other position (b).

a. Find the expression for i(t) at the position (b), b. What is the initial voltage across the inductor just after the switch has

been moved to position (b), c. Does this initial voltage make sense in terms of circuit behavior? d. How milliseconds after the switch has been moved does the inductor

voltage equal 24 V? e. Plot v(t) versus t.

Solution:

a. The switch has been for a long time in position (a), so the inductor is a short circuit across the 8A current source. Therefore the inductor carries an initial current of 8 A. This current is oriented opposite to the reference direction for i; thus I0 is 8 A. When the switch is in position (b), the final value of i will be 24/2=12 A. The time constant of the circuit is 200/2 or 100 ms. Then :

Aeei tt 101.0/ 2012)128(12 =+= b. The voltage across the inductor is shown in the attached figure as :

Veedt

diLv tt 1010 40)200(2.0 ===

c. Yes, in the instant after the switch has been moved to position (b), the inductor sustains a current 8 A which causes a 16 V drop across the 2 resistor. This voltage drop adds to the drop across the source, producing a 40 V drop across the inductor.

d. Solving the expression: te 104024 = then t=51.08 ms.

V

t

ab

+

-

i v

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Example 5.3: The switch in the circuit has been in position x for a long time. At t=0, the switch moves instantaneously to position y. Find:

1. vc(t) for 0t , 2. vo(t) for + 0t ,

3. io(t) for + 0t and 4. The total energy dissipated in the 60 K resistor.

Solution:

1. Because the switch has been in position x for a long time, the capacitor will charge to 100V and be positive at the upper terminal. Replace the resistive network connected to the capacitor at t=0+ with a single 80 K resistor. Hence the time constant of the circuit is )1080)(105.( 36 o , or 40 ms. Then:

Vetv tC25100)( = , 0t .

2. The easiest way to find vo(t) is to note the resistive circuit forms a voltage divider across the terminals of the capacitor. Thus

Vetvtv tCo2560)(

80

48)( == , + 0t .

3. From s law:

mAetv

ti to25

30

1060

)()( =

= , + 0t .

4. The power dissipated in the 60 K resistor is mWetitp to

5032 60)1060)(()( == , + 0t . The total energy dissipated is

==0

32 2.1)1060)(( mJdttiw o

t

)(

Example 5.4:

The switch has been in position (1) for a long time. At t=0, the switch moves to position (2). Find:

a) vo(t) for 0t and b) io(t) for + 0t .

Solution:

a) The initial value of vo is 40(60/80) = 30 V. Find the Norton equivalent with respect to the terminals of the capacitor for

0t , by first computing the open-circuit voltage, which is given by 75 V source divided across the 40 K and 160 K resistors:

VVoc 60)75(10)16040(

101603

3

=+

=

Next, calculate Thevenin resistance, as seen to the right of the capacitor, by shorting the 75 V source and making series and parallel combinations of the resistors: RTH = 8000 + 40,000 || 160,000 = 40 k The value of the Norton current source is the ratio of the open-circuit voltage to the Thevenin resistance, or 60/(40 x 103) = -1.5 mA:

Noting that vo(0) = 30 V, IsR = -60 V and RC = 10 ms, so the solution for vo is: [ ] 0,9060)60(3060 100100 +=+= tVeev tto

b) Noting that Is = -1.5 mA and vo/R = (30/40) x 10-3 or 0.75 mA:

+ = 0,25.2 100 tmAei to

1 2

+

-Vo

t0