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Elg2138 HW#7 Solution

P10.5-2

Apply KCL at node a, using (current out)=0, we have

4 cos 20.25 0

1

v t d v i

dt 

(1) 

The first term on the left comes from Ohm’s Law and the second term comes from the vi-characteristic of the capacitor.

Apply KVL to the right mesh, using (voltage drop)=0 cw direction, we have

4 4 0 4 4d d 

i i v v i idt dt  

(2)

Substitute (2) in (1), we have

 Rearranging, we have

2

25 5 4 cos 2

d d i i i t  

dt dt  

(3) 

Now use 2Re{ }

 j t mi I e

  and 2

4 cos 2 4 Re{ } j t 

t e to write

2

2 2 2 2

2Re{ } 5 Re{ } 5 Re{ } 4Re{ }

  j t j t j t   j t 

m m m

d d   I e I e I e e

dt dt  

   

2

2 2 2 2

2Re 5 5 Re{4 }

  j t j t j t   j t 

m m m

d d   I e I e I e e

dt dt  

 

 

2 2 2 2Re 4 5 2 5 Re{4 }

  j j t j j t j j t j t  

m m me I e j e I e e I e e

   

 

4 5 2 5 4  j j j

m m me I j e I e I      

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4 4 4

0.398 844 5 2 5 1 10 10.05 84

 j

m I e j j

   

Returning to the time domain, we write

}

  0.398 cos 2 85 Ai t t   

P10.8-16Represent the circuit in the frequency domain using impedances and phasors.

We have =20. Doing series reduction of each of the 3 branches, we have

For the top branch,

1

2

3

125 20 2 25 15 29.2 31

20 0.002

120 20 10 22.36 26.6

20 0.005

40 20 2 40 40 56.57 45

 j j j

 j j

 j j

Z

Z

Zfor the right branch

 

Perform parallel reduction on the two vertical branches, we have                        

=

  

   = 18.665 – j2.676

By the generalized Ohm’s Law, the current I coming out of the voltage source V is IT = V / ZT 

where V =      and

ZT = 43.665 + j12.324 = 45.37      

By current divider, we have =                        =      

Converting to the time domain:

I

 

0.118cos 20 6.1 Ai t t   

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P10.9-10

Mesh 1 : using KVL , drop = rise, CW direction

=1030 

  =1030 (1)

Mesh 2 : using KVL , drop = 0, CW direction  

  =0

Mesh 3 : using KVL , drop = 0, CW direction

 

 

=0

   

Let

 

 

   

  

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 = (1030) (  where =tan

-18/2 =75.964 

= 82.462105.964 

Using Cramer’s rule yields 

 Then   v(t) = Re{

t+44.0 V 

P10.10-5

First determine ocV using mesh current method.

Mesh 1 : using KVL , drop = rise, CW direction,

1 1 2 1 2600 300( ) 9 (600 300) 300 9 0  j j j I I I I I (1) 

Mesh 2: using KVL , (voltage drop) = 0, CW direction, (2)

Since by Generalized Ohm’s Law,  (3)

Substituting (3) in (2), we have  (4)Using Cramer’s rule (see example in P10.9.10) 

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2 0.0124 16 A I  

Then

oc 2300 3.71 16 V V I  

Next, we determinescI :

Mesh 2: using KVL , (voltage drop) = 0, CW direction,

     Since    

Mesh 1: using KVL , drop = rise, CW direction,    

Therefore,   

    

The Thevenin impedance is therefore,

ocT

sc

3.545 16247 16

0.015 0

VZ

I 

The Thevenin equivalent is

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P10.10-9

1

( 3)(4)2.4 53.1

3 4

=1.44 1.92

 j

 j

 j

Z 

2 1

4

1.44 2.08

2.53 55.3

 j j

Z Z

 

33.51 37.9

2.77 2.16 j

Z

 

By current divider,       

3.51 37.93.51 37.92.85 78.4 2.85 78.4 1.9 92 A

2.77 2.16 2 5.24 24.4 j

I