solutions to assignment 9

7/23/2019 Solutions to Assignment 9

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Solutions to Assignment 9

1. 1.13cos(2t+45) V

Since the frequency of the source is a constant 2 rad/s, replace each element with

its equivalent impedance:

The 0.5F Capacitor is replaced with an impedance of value 1/jC= 1/j(2*0.5)=-j;

The 1H Inductor is replaced with an impedance of value jL= j(2*1)=2j;

Now, we use superposition and Voltage division rules to find the Voltage across

the central 0.5F capacitor.

From V1 alone, the voltage across the 1H inductor is :

VL= V1*(6j/6j-j);

The voltage across the 0.5F capacitor = VL(-j/-j-2j)= VL/3 = (2*V1)/5.

By symmetry, the voltage across the capacitor due to V2 is (2*V2)/5 in the

opposite direction.

Net Voltage = 0.4*(V1-V2) = 0.4 (2- (-2j))=0.8(1+j)


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This can be expressed as 0.8*sqrt(2)*cos(2t+45).

2. 8cos(t)/5

Assume that the mesh currents in the 2 loops are I1 and I2 respectively (clockwise

direction).

The 0.5F Capacitor is replaced with an impedance of value 1/jC= 1/j(2*0.5)=-j;

The Inductor is replaced with an impedance of value jL= j(2*1)=2j;

The phasor equations are :

V1= jL

1*I

1+ jM*I

2+ (I

1-I

2)/jC; and

(I1-I2)/jC= jL2*I2+ jM*I1

Substituting values, we have :

2= jI1 + 0.25j*I2+ (I1-I2)/j; and


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(I1-I2)/j= j*I2+ 0.25j*I1

Solving, we have I1=0, I2=-1.6j

V1=1.6j/j=1.6 ; which implies 1.6cos(t)

3.108cos(1000000t + 37.87) V

As a first step, we convert the current source is parallel with the 10ohm resistorinto a voltage source V2= I1*10 =100exp(j0); with a series resistor R2=10

It is given that the frequency of the circuit is 106

rad/s.

The 0.1F Capacitor is replaced with an impedance of value 1/jC= 1/0.1j=-10j;

The 5H Inductor is replaced with an impedance of value jL= 5j;

The parallel combination of resistor and capacitor has an equivalent inductance

of (5*-10j)/ (5-10j) = 4-2j

If we introduce mesh currents Ix and Iy in the clockwise direction, we have:

Now, we can write the voltage equations as follows :

V1-Va =Ix*(4-2j);


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Va-V2= Iy(10-10j);

Va= jwL*(Ix-Iy)=5j(Ix-Iy).

We have 3 equations and 3 unknows (Va,Ix and Iy). Solving, we get the answer.

4. A=1/3, Phi = -90 degrees

Assume that the mesh currents in the 2 loops are I1 and I2 respectively (clockwise

direction). Given the frequency of the circuit is 2rad/s ( from value of V3), we

have : jC= j(2*0.5)= j;

The phasor equations are :

V1=I1+(I1-I2)/jC;

(I1-I2)/jC=I2+V2;

V2=I2/jC.

Solving these, we have :

V1/V2= 3j;

V2/V1= -(1/3)j . Hence the solution.


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5. 1/sqrt(LC)

Given RC=RL; and |IC|=|IL|.

ZC=(RC+1/jC) and ZL= (RL+jL).

Since the Voltage is the same, the 2nd

condition (on the currents) implies :

|ZC|=|ZL|

Squaring we have

|ZC|2=|ZL|

2

RC2+(1/C)

2= RL

2+(L)

2.

Since RC=RL, 4=1/(LC)2

=1/sqrt(LC)

5. 14.19


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To find the effective resistance, short the voltage source and find the effective

resistance across the terminals of RL

Reff=[ ( (10 || 15)+ 20) ||5]+10

Reff=* (620)|| 5+ 10

Reff = *(265/31)+10 = 14.19

7. 1 Ohm


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Here , the only the frequency of the voltage source is important. Since the

frequency of the source is a constant 2 rad/s, replace each element with its

equivalent impedance:

The 0.25F Capacitor is replaced with an impedance of 1/jC= 1/j(2*0.25)=-2j;

The 0.5H Inductor is replaced with an impedance of value jL= j(2*0.5)=j;

The 2 capacitors are in parallel (when viewed from the terminals of Zx), So the

effective value of this impedance is -2j*-2j/-4j = -j. This is in series with a 1

resistor, so the effective impedance of this branch is (1-j).

Also, the value of the impedance due to the resistor and inductor in series with

the Voltage source V1 is (1j)

Therefore, the effective impedance Zeff, as seenfromZx is

(1+j)*(1-j)/ (1+j +1-j)= 1.

The maximum power is transmitted when Zx= (Zeff) . Here, since Zeff=1,Zx=1.


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8. 75V, 10kOhms

Assume that the Thevenin equivalent voltage is V1 and Thevenin equivalent

resistance is R1 for the given circuit. In the first case, by voltage division, we

have:

V1(15k)/ (15k R1) = 45V . Eqn1

And the 2nd experiment gives:

V1(5k)/ (5k R1) = 25V. Eqn2

Dividing equation 1 by 2, we have :

3*(5k + R1)/(15k R1) = 45/25 =9/5.

This gives R1=10k .

Substituting in one of the equations, we have, V1= 75V

9. 86.4 V, 43.2kOhms


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10. 35 Ohms, 315W


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We first find the Thevenin equivalent with respect to the terminals of RL.

Assume the current entering the coil on the left is I1 and the voltage across the

coil is V1. Similarly, the voltage across the 2nd

coil (+ve terminal at the dot) and

current flowing to the load through the transformer be I2

Given that the transformer is ideal, we have:

V2=(1/4)V1 and I1=-(1/4)I2;

For the open circuit case:

I2=0; hence I1=0;

V1=840exp(j0)V ; V2= 210exp(j0)V.

The thevenin voltage VTh in this open circuit case isV2;

VTh= -210exp(j0).

For the short circuit case (RL is removed and the terminals are shorted):


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The mesh equations (given currents I1 and I2) are:

840exp(j0)= 80I1-20I2+V1;

0 =20I2-20I1+V2.

The mesh equations, combined with the constraint equations for the ideal

transformer gives:

I2=-6A.

The thevenin resistance RTh = VTh/I2 = -210/-6=35.

Thus, maximum power is delivered when RL=RTh = 35

The value of this maximum power is :

Pmax= I2R=(V/(RTh+RL))

2*RL= 315W

11. 80 Ohms


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12. |Zab| = 56.91, arg(Zab) = 13.28 degrees

From the above, we have :

ZAB= (Vg/Ig) 10 = (340/(5-j)) 10 = 55.38+j13.08

Hence the answer.

NOTE : Some of the diagrams/solutions in the assignment are obtained from the

book Electric Circuits, 8th

Edition by Nilsson and Riedel.

solutions to assignment 9

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