Download pdf - Circuit Theory - Solved Assignments - Spring 2005

Assignment 1 (Spring 2005)

(Solution)

CIRCUIT THEORY (PHY301) MARKS: 30

Q.1. Sol.

Figure 1

Using Figure 1: Calculate the voltage drop and current for each resistor.

Step #1: Reduce the circuit to one equivalent resistance value. R and R1 2 are connected directly in parallel and can be replaced with an equivalent resistor, RX. The resulting circuit is shown in Figure A.

Figure A

RX and R4 are in series (sum these two values), therefore, they can be combined to produce value RY.

RY = RX + R4 = 10 + 6 = 16

Figure B shows the resulting circuit.

Figure B

RY and R3 are now directly in parallel and can be replaced with an equivalent resistor, RZ.

The resulting circuit now looks like the one in Figure C:

Figure C

The total equivalent resistance is therefore the sum of RZ and R5.

R = R + R = 12 + 3 = 15 T Z 5

Step #2: Calculate total current in the circuit using Ohm’s Law.

Step #3: Determine the voltage drop of each resistor. Figure D shows the current flow in the circuit (indicated by the arrows).

Figure D

The current leaving the supply (+) is the total current entering the circuit, and the current leaving R5 is also the total current. Therefore, the current through resistor R5 is the only known current (which is the same as the total current, 8 amps). Using Ohm's Law, you can calculate the voltage drop across R5.

V = I × R = 8 A × 3 R5 R5 5 = 24 V

In the above Figure , the resultant resistor RZ would have a voltage drop. If this voltage drop were added to the voltage drop of R5, the result would be equal to the applied voltage (as per Kirchhoff’s Voltage Law). Therefore: E = VRZ + VR5 By transposing the formula, we can find the value of VRZ.

E = VRZ + VR5 OR VRZ = E – VR5 VRZ = 120 – 24 = 96 V

Or we can find it out by voltage division formula

V12= (12x 120)/ 12 +3 = 96V

Figure E shows the comparison of RZ to a section of the series-parallel circuit that we simplified to find RZ. By studying Figure E, we can see that the voltage across RZ is 96 volts. Also, we can see that R3 is directly in parallel with RZ; therefore, R3 would have the same voltage across it as RZ (96 V). To find the voltage drop across each resistor, we must know the resistance and the current through each resistor. We know the voltage across R3 and the resistance of R3; therefore, we can calculate the current through R3.

Figure E

Figure F

From Figure (F) by using voltage division formula

V4 = (6 x 96)/ 10 +6 = 36V IR4= VR4/ R4 = 6A

We have now found the voltage drop across R3, R4, and R5; also, the sum of these three voltages equals the supply voltage (E). Figure G shows the circuit with the known voltages and their location.

Figure G

If we now analyze the circuit, we can make certain observations:

1. The voltage of R1 is equal to the voltage of R2. 2. The voltage of R1 added to the voltage of R4 is equal to the voltage of R3. 3. If voltage of R4 is subtracted from the voltage of R3 we get the voltage of R1

(and the voltage of R2 since they are equal in value).

Therefore, we can now calculate the voltage of R1 and R2:

VR1 = VR2 = VR3 – VR4 VR1 = VR2 = 96 V – 36 V VR1 = VR2 = 60 V

All of the voltage drops across each resistor have now been determined. The current through each resistor has been found except for IR1 and IR2. Step #4: Find IR1 and IR2.

Q.2. (a) Consider the circuit shown below. Suppose that V=30V, R1=10KΩ , and VO =10V. Find R2 and also the power dissipated in R2. (b) Consider the circuit shown below. Suppose that V=12V, R2=140 Ω , and VO =8.5V.Find R1 and also the power dissipated in R1. Sol. (a)

Vo = ( R2)x (V)/(R1 + R2)

10 = 30 x R2 / 10 + R2

3R2= 10 + R2

R2 = 5 kΩ Po = Vo

2 / R2 = 100/5 = 0.02W=20mW (b)

Vo = ( R2)x (V)/(R1 + R2)

8.5 = 140 x 12 / 140 + R1

1680 = 1190 + 8.5R1

R1 = 57.6 Ω P1 = Vo

2 / R1 = (3.5)2/57.6 = 0.212W

P2 = Vo2 / R1 = (8.5)2/140 = 0.52W

------ Good Luck -----

Assignment 2(Spring 2005)

(Solution)


Due Date: 14/04/2005

Q.1.

Use nodal analysis to find VO in the given network. Identify and label each node otherwise you will lose your marks.

Sol. First, we will labeled the diagram, Now we write KCL equation for node “V1”, V1 / 5Ω + (V1 – 3) / 1 Ω + (V1 - 4VO )/5= 0 ----- (1) From the figure it is clear that VO = 2(V1/5) --------- (2) Substitute the value of Vo in the equation (1)

V1 / 5 + (V1 – 3) / 1 + (V1 – (4(2(V1 /5))/5= 0 V1 / 5 + (V1 – 3) / 1 - 3V1 / 25= 0

After simplifying V1 = 75/27

Put this value in equation 2 we have

VO = 75/27 x 2/5 VO = 1.11 Volts

Q.2. Use nodal analysis to find voltage at each node also finds current IO as indicated in the given network. Identify and label each node otherwise you will lose your marks.

Sol. First, we will label the diagram and identify each node, In the above network we found that there is a voltage between two non reference nodes, so we will consider it a super node, we will redraw the figure as Now Constrain Equation will be V3 – V2 = 50V (1) KCL equation at Node V1 will be,

(V1 – V2)/5Ω + V1/200Ω + (V1-V3)/100Ω = 0 OR V1+ 40V1 - 40V2+ 2V1 – 2V3= 0 43V1 - 40V2 – 2V3= 0 (2)

Now we will write the equation for super node,

(V2 – V1)/5Ω + 0.2+ V3/50Ω + (V3-V1)/100Ω = 0 OR

(20V2- 20V1 + 20 + V3- V1 + 2V3)/ 100 = 0 20V2- 20V1 + V3- V1 + 2V3 = - 20 - 21V1 + 20V2 + 3V3 = - 20 (3)

After solving equation (1) (2) and (3) simultaneously we have V1 = -45.23V, V2 = -48.69V and V3 = 1.31V

We have at V2 (V2-V1) /5 + 0.2 + I0 = 0

I0 =– 0.2A+ (V1 – V2) /5 =-0.2 + (-45.23 +48.69)/5 I0 = 0.494A

------ Good Luck -----

Assignment 3(Spring 2005) (Solution)


Due Date: 21/04/2005

Q.1.

Use nodal analysis to find Voltage at each node in the given network. Identify and label each node otherwise you will lose your marks.

Sol. First, we will labeled the diagram,

At Node V1

(V1 + 3Vo –V0)/2K + (V1 –Vo)/4k = 3 Multiply both sides by 4 we have (2V1 + 6Vo –2V0) + V1 - Vo = 12 3V1 + 3Vo =12 ----- (a) At Node Vo

Vo/1k + (Vo-V1) /4k +(Vo - 3Vo –V1)/2K=0 Multiply both sides by 4 we have 4Vo + Vo –V1 + 2Vo -6Vo – 2V1 =0 -3V1 +Vo = 0 ------ (b) Solving (a) and (b) simultaneously we have, V1= 1V, Vo = 3V

Q.2. You are given the network below. Use any method you desire to find I and Vab. Show your complete work. Label circuit diagram properly.

Sol. First, we will labeled the diagram,

We will start from point A, -30 + 3I – 10 +5I + 8 = 0 8I = 32 I = 4A --------- (1) Now for Vab start at point “b” we have -Vab + 5I + 8=0 ------- (2) Put the value of I from (1) in (2) we have Vab = 28 V

Q.3.

Use Mesh analysis to find Current IO in the given network. Identify and label each mesh otherwise you will lose your marks.

Sol. First label each mesh current

For Mesh I:

First, start with I1. It passes through 10 ohm resistor and drops 6 V due to a voltage source Mesh current I1 opposes I2 in the 2 ohm resistor,

These yields the equation

10 I1 -6V + 2 (I1 - I2) = 0

12 I1 -2I2 + 0I3 = 6 -------- (A)

For Mesh II:

Mesh current I2 passes through 4 ohm resistor, opposes I1 in the 2 ohm resistor, opposes I3 in 1 ohm resistor, and drops 8 V due to a voltage source.

4I2 + 2 (I2 - I1) + 1(I2 – I3) + 8V= 0

-2I1 + 7I2 - I3 = -8 -------- (B)

For Mesh III:

Mesh current I3 passes through 5 ohm resistor, drops 8 V due to a voltage source, opposes I2 in the 1 ohm resistor and drops 6 V due to a voltage source.

5I3 -8V + 1(I3 – I2) + 6V= 0

0I1 - I2 + 6I3 = 2 -------- (C)

Solving (A)(B) and (C) simultaneously we have,

I1=0.329 A, I2 = -1.0256, I3= 0.162

I0 = I3 –I2

I0= (0.162 – (-1.0256))

I0= 1.188A

------ Good Luck -----



Due Date: 04/05/2005

Q.1. Use Mesh analysis to find Current through each mesh in the given network also find out the ratio V2/V1. Identify and label each mesh and also show each step of calculation otherwise you will lose your marks.

Sol.

First, we will label the diagram

Mesh I

3I1 -2I2 = V1 ------- (A) Mesh II -2I1 + 10I2 -4I3 -2V = 0 ----- (I) But we know that V= 2(I1-I2) Put this value in (I) we have -2I1 + 10I2 -4I3 -2(2(I1-I2)) = 0 -6I1 + 14I2 - 4I3 = 0 ------ (B) Mesh III -4 I2 + 8 I3 +2V = 0 ------ (II) By Putting V= 2(I1-I2) in (II)

4I1 - 8I2 + 8I3 = 0 ------ (C) Solving (A) (B) and (C) simultaneously we have I1 = 0.4545 V1

I2 = 0.1818 V1

I3 = - 0.0454 V1

We know that V2 =4I3 ---- (III) Put I3 = - 0.0454 V1 in (III) we have V2 =4 (- 0.0454 V1) V2 / V1 = -0.1816

Q.2. Use Mesh analysis to find Current through each mesh in the given network also find the voltage V1. Identify and label each mesh and also show each step of calculation otherwise you will lose your marks. Sol. First, we will labeled the diagram Mesh I

50I1 -30I2 = 3V ------- (A) Mesh II -30I1 + 40I2 + V1 = 0 ----- (B) Mesh III 50 I3 – V1 = 0 ------ (C) Constraint equation I 3 – I 2 = 0.5A ------ (D) Solving (A) (B) (C) and (D) simultaneously we have

I1 = - 0.13A

I2 = - 0.32A I3 = - 0.178 A V1 = 8.89V

------ Good Luck -----



Due Date: 20/06/2005

Q.1. Identify and label each mesh. Draw and label the circuit diagram, otherwise you will lose your marks. Write each step of calculation and also mention the units of each derived values.

(a) Find current through all meshes using mesh analysis. Solve equations by using matrices. (b) Find the magnitude of current through 7Ω. (c) Is the direction of the current through the 7-Ω resistor up or down?

Sol. First, we will label the diagram

Mesh I 3I1 + (I1-I2) +4V + 5(I1-I2) – 6V= 0

9I1- 6 I2 + 0I3= 2V ------- (A) Mesh II 3V + 2I2 + 5(I2-I1) – 4 +1(I2-I1) = 0 -6I1+8 I2 + 0I3= 1V ----------- (B) Mesh III 7I3 +7 V = 0 ------ (C) In matrix form

1

2

3

9 6 0 26 8 0 10 0 7 7

III

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Where,

9 6 06 8 00 0 7

9(56) 6( 42)504 252252

A−

= −

= + −= −=

1

217

A =−

and

2

9 2 06 1 00 7 7

2529(7) 2( 42)

25263 84

2520.5833

I

A

−−

=

− −=

+=

=

Where, and 2

9 2 06 1 00 7 7

A = −−

3

9 6 26 8 10 0 7

2529( 56) 6(42) 0

252504 252

2521

I

A

−−

−=

− + +=

− +=

= −

Where, and

3

9 6 26 8 10 0 7

A−

= −−

(b) I 7Ω =-1 (c) Actual direction is in opposite to the assumed. i.e. in upward.

Q.2. Using superposition find a) the current through the 2-Ω resistor due to the voltage source b) the current through the 2-Ω resistor due to the current source c) the power delivered to the 2-Ω resistor

Draw and label the circuit diagram, otherwise you will lose your marks. Write each step of calculation and also mention the units of each derived values.

Sol.

a) open the current source

⇒ Fig (1)

From fig (1) we have I2Ω= (8)(Is)/8+2 = 0.8 A

I1Ω= (3)(Is)/(3+1+5)

= 3/9=1/3 A I2Ω= 1/2(I1Ω) = 1/6 A

I2Ω total = 0.8-1/6 = 0.633 P = 0.633x2 = 0.8W

Q.3. Using source transformation finds IO in the following network. Draw and label each circuit diagram, otherwise you will lose your marks. Write each step of calculation and also mention the units of each derived values.

Sol. 6V source is in series with 12KΩ, we will convert it into current source of value = 6V/12kΩ = 0.5mA In modified circuit 0.5mA current source and 12KΩ resistor will be in parallel.

------ Good Luck -----



Due Date: 27/06/2005

Q.1. Find Rth (Thevenin’s resistance ), Vth (Thevenin’s Voltage) and IN (Norton’s current) in the circuit below. Draw and label the circuit diagram, otherwise you will lose your marks. Write each step of calculation and also mention the units of each derived values.

Sol. To find RTH we will short the voltage source and open the current source.

RTH =4Ω

1A current source is in parallel with 6Ω, we will convert it into voltage source

of value = 1A x6 Ω = 6V

In modified circuit 6V voltage source and 6Ω resistor will be in series.

Q.2. a) Determine the short-circuit current, that is, the current flowing through a wire with zero ohm

connected between a-b. b) Determine the open-circuit voltage, that is, the voltage between a-b with the short removed.

c) Determine VTH (Thevenin’s Voltage) and RTH (Thevenin’s resistance) looking into a-b. d) Draw the Thevenin’s circuit. e) Using your Thevenin circuit, connect a 25Ω resistor between terminals a-b and determine the

voltage, Vab. Draw and label the circuit diagram, otherwise you will lose your marks. Write each step of calculation and also mention the units of each derived values.

Sol.

------ Good Luck -----



Due Date: 10/07/2005

Q.1. (a) Find the Thevenin’s equivalent voltage between the nodes A and B.

(b) Find the Thevenin’s equivalent resistance between the nodes A and B.

(c) Draw the resulting equivalent circuit.

(d) Using the constant voltage drop model for the diode D1 (With VD = 0.7V), estimate the

diode current.

Sol. (a) Thevenin’s equivalent Voltage between the nodes A and B

Vth = 5V (5k/5k+1k) = 4.167V

(b) Thevenin’s equivalent Resistance between the nodes A and B Rth = 2k + (5k x1k)/1k+5k = 2.833kΩ

(c) Resulting equivalent circuit

(d) Diode current I = (4.167-0.7)/2.833k = 1.224mA Q.2. A 1-mA diode (i.e., one that has VD =0.7V at iD = 1mA) is connected in series with a 200-Ω resistor to a 1.0V supply.

(a) Provide a rough estimate of the diode current you would expect.

(b) If the diode is characterized by n=2, estimate the diode current more closely using iterative analysis.

Sol. First we will draw the circuit,

(a) Rough estimate I= (1-0.7)/200 = 1.5mA= 0.0015A (b) Iteration:

We know that I = Isev/nVT

Where I= 1mA and V =0.7 , n = 2 and VT= 0.025

So Is = 8.31 x10-10 A = 0.000000000831A

Step I: V= nVT ln(I/IS) =(2)(0.025) ln(0.0015/ 8.31 x 10-10) = (0.05) ln (1805054.152)

= 0.05 x 14.40 = 0.72V --- (A) I = 1V- 0.72V/200Ω = 0.0014 A Step II: V= VT ln(I/IS) =(2)(0.025) ln(0.0014/ 8.31 x 10-10) = (0.05) (14 .33) = 0.72V ---(B) The value of (A) and (B) are same so we will stop iteration. Q.3. Determine the dc load voltage for the circuit shown below.

Primary voltage is in rms so, we can calculate the peak voltage V

1(pk) = V

rms/0.707

= 89/0.707 = 125.88V

pk Primary voltage and turn ratio is kno

V2(pk)

= =(125.88) x 3/7

wn so we can determine the secondary voltage (N2/N1)V

1pk

=53.95Vpk

Now load voltage can be calculated as V

L(pk) = V

2(pk) /2 – 0.7

= (53.95/2) -0.7 = 26.275V

pk Now dc value of the voltage can be calculated as

Vave

= 2VL(pk)

/ Π

= 52.55/ Π = 16.72 Vdc

------ Good Luck -----

Q.1. A single measurement indicates the emitter voltage of the transistor in the circuit given below to be 1.0V. Under the assumption that |VBE| = 0.7V, what are VB,IB,IE,IC,VC,β,α ?

Solution We know that, 1) IE = (5-VE)/5

= 4/5 = 0.8mA

2) IB =VB/RB = 0.3/20 = 15 µA 3) IE *RE+VEB+VB= 5

0.8*5+0.7+ VB = 5 VB = 0.3V

4) IC=βIB

= 52.33*15 µΑ = 0.785mΑ 5) VC = IC*5-5 = 0.785mA*5-5 = -1.075V 6) We know that,

IE= (β+1)ΙΒ

β+1= IE/ ΙΒ

β +1 = 53.33 β = 53.33−1 = 52.33 7) IC= α IE

α = IC/ IE = 0.785mA /0.8mA = 0.98125