Circuit Theory - Solved Assignments - Semester Fall 2004

Assignment 1 (Fall 2004) Solution

CIRCUIT THEORY (PHY301) MARKS: 30

Due Date: 19/10/2004

Q.1. Compute the power that is absorbed or supplied by the elements in the network given below.

Sol.

Power absorbed or supplied by each element P36V = VI= (36) (-4) = -144 W

P1 = VIx= (12) (4) = 48 W P2 = VI= (24) (2) = 48 W PDS = VI= (4) (-2) = -8 W P3 = VI= (28) (2) = 56 W

Q.2. An automobile uses two halogen headlights connected as shown in the circuit. Determine the power supplied by the battery if each headlight draws 3A of current. Sol.

PS = IV= (3+3) (12) =72 W

Circuit Theory - Solved Assignments - Semester Fall 2004

Q.3. Find Req for the following circuit. Draw and label the circuit diagram of each step otherwise you will lose your marks because diagram of each step contains marks.

Sol.

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



Due Date: 01/11/2004

Q.1. Find IO in the circuit below. Draw and label the circuit diagram of each step otherwise you will lose your marks because diagram of each step contains marks.

Sol.

First of all we will find out the total current Ix in the network then find out the current Io through 4kΩ.

Q.2. Use nodal analysis to find VO in the given network. In addition, determine all branch currents and check KCL at every node. Identify and label each node otherwise you will lose your marks. Label circuit diagram properly.

Sol.

KCL equation at super node is

(V1-6)/10 + V1/10 + IX + V2/20 = 0 IX = (6 - V1)/10k ----------------- (I) (V1-6)/10 + V1/10 + (6 - V1)/10+ V2/20 = 0

After simplifying we have 2V1 –V2 = 0 ---------------- (A)

Node 1 and 2 constitute a super node Constraint equation will be

V1 – V2 = -12 -------------------- (B) After solving equation (A) and (B), we have

V1 = - 4V V2 = 8V

By voltage division rule we have Vo = (10k) (V2)/ (10k+10k) Vo = 4V IX = (6 - V1)/10k = (6+4) / 10k IX = 1mA I1 = V1/10k = -4/10

I1 = - 0.4mA From the above figure it is clear that I2 = IX = 1mA

I3 = V2 /20k = 8/20k

I3 = 0.4mA At Node V1

IX = IY + I1IY = IX - I1 = 1mA-(-0.4mA) IY = 1.4mA

Q.3. In the network given below the power absorbed by the 4Ω resistance is 100 W. Find VS.

Sol.

P=100 W= IA

2 x 4 So,

IA= 5A ----- (A) Where by ohm’s Law

VX = 5 x 4 =20 V IB = 20 /10 IB = 2A ------ (B)

From (A) and (B) we have IS = IA + IB = 7A VS = 3 x 7 + VX = 21 + 20

VS = 41 V

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



Due Date: 10/11/2004

Q.1. Use nodal analysis

(i) Find the current I. (ii) Find the power absorbed by the 50 ohm resistance. (iii) Find the power supplied by the 20 V source.

Identify and label each node otherwise you will lose your marks. Label circuit diagram properly.

Sol.

(VA-20)/10 + (VA-30)/40 + VA/ 70 = 0 After simplifying we have VA = 770/39 VA =19.74 Volts --------- (A) (a) I = (VA -20)/10 = (19.74 - 20) /10 = -0.026 A (b) By voltage division rule we have V50 = (VA) (50) /( 50 +20) = (19.74) (50) / (50 +20) V50 =14.1 P50 = V2

50/ 50 = (14.1)2/50 =3.98 W (c) PSUP(20) = -I x 20 = - (-0.026)20 = 0.52W

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

Sol.

From the above figure we have -40-60+50 + 50I = 0 50 I = 50 I=1A ---- (A) 8I -40 +15I+ 12I +VCE= 0 Put the value of I from (A) VCE = 40 – I (35) VCE = 5V Q.3. Identify and label each mesh. Find current through all meshes using mesh analysis for the following network. Draw and label the circuit diagram if necessary, otherwise you will lose your marks. Sol. Constraint equation will be I1 –I3 =3 ------- (A) Mesh 1 10(I1-I2) + VS + 20(I1-I3) = 20

30I1-10I2 + VS - 20I3 = 20 --------------(B) Mesh 2

- 10I1+ 20I2 = 0 ---------- (C)

Mesh 3

-20I1 + 26I3 -VS = 0 ------------ (D) Solving (A) (B) (C) and (D) simultaneously we have I1= 3.45A, I2= 1.73A I3 = 0.45A ,VS= -57.27

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



Due Date: 24/11/2004

Q.1. Identify and label each mesh. Find current through all meshes using mesh analysis also find current I through 6 ohm resistance and voltage across 4 ohm resistance. Solve equations by using matrices. . Draw and label the circuit diagram if necessary, otherwise you will lose your marks.

Sol.

First label each mesh current

This problem is simplified by the fact that there are two current sources which remove two of the unknowns.

• I3 = -3A • I4 = 1A

For Mesh I:

Now we are left with two unknowns. First, start with I1. It passes through a 2 ohm resistor, opposes I2 in a 6 ohm resistor, opposes I3 in a 4 ohm resistor, and opposes I4 in a 6 ohm resistor. These yields the equation

2 I1 + 6 (I1 - I2) + 4 (I1 - I3) + 6 (I1 - I4) = 0

2 I1 + 6 (I1 - I2) + 4 (I1 + 3) + 6 (I1 - 1) = 0

18 I1 - 6 I2 = -6 ---------- (A)

For Mesh II:

Mesh current I 2 opposes I 1 in the 6 ohm resistor, I 3 in the 3 ohm resistor, and drops 12 V due to a voltage source

6 (I2 - I1) + 3 (I2 - I3) = -12

6 (I2 - I1) + 3 (I2 + 3) = -12

-6 I 1 + 9 I2 = -21 ------------ (B)

Put (A) and (B) into a matrix gives

1

2

18 6 66 9 21

II

− −⎡ ⎤⎡ ⎤ ⎡=⎢ ⎥⎢ ⎥ ⎢− −⎣ ⎦ ⎣⎣ ⎦

⎤⎥⎦

Where

18 6

6 9D

−⎡ ⎤= ⎢ ⎥−⎣ ⎦

D = (18)(9) - (-6)(-6) = 162 - 36 = 126

1

6 621 9

ID

− −⎡ ⎤⎢ ⎥−⎣ ⎦=

I1 = ( (-6)(9) - (-21)(-6) ) / 126 = (-54 - 126) / 126 = -180 / 126 = -10/7

2

18 66 21

ID

−⎡ ⎤⎢ ⎥−⎣ ⎦=

I2 = ( (18)(-21) - (-6)(-6) ) / 126 = (-378 - 36) / 126 = -414 / 126 = -23/7

The mesh currents are:

I1 = -10/7 A

I2 = -23/7

I3 = -3 A

I4 = 1 A

From the currents, I and V can be found

I = I1 - I2 = (-10/7) - (-23/7)

I = 13/7 A

V = 4 (I1 - I3) = 4 (-10/7 + 3)

V = 44/7 V

Q.2. Identify and label each mesh. Find current through all meshes using mesh analysis also find current through each resistance. Solve equations by using matrices. Draw and label the circuit diagram, otherwise you will lose your marks.

Sol.

First label each mesh current

For Mesh I:

First, start with I1. It passes through a 9 k ohm resistor, opposes I3 in a 10k ohm resistor, . and drops 50 V and 100 V due to a voltage source

These yields the equation

9 I1 + 10 (I1 + I3) + 100 +50 = 0

19 I1 +0I2 + 10I3 = -150 -------- (A)

For Mesh II:

Mesh current I 2 opposes I 3 in the 3 k ohm resistor, I2 passes through a 4 k ohm resistor, and drops 50 V due to a voltage source


3(I2 – I3) + 4 I2 + 50 = 0

0 I1 + 7I2 - 3I3 = -50 -------- (B)

For Mesh III:

Mesh current I 3 and I 1 passes in the same direction through 10 k ohm resistor, Mesh current I 3 opposes I 2 and I3 passes through 7 k ohm resistor


10(I3 + I1) + 3(I3 – I2) + 7I3 = 0

10 I1 - 3I2 +20I3 = 0 -------- (C)

We will write equation (A) , (B) and (C) in matrix form

1

2

3

19 0 10 1500 7 3 5010 3 20 0

III

−⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥− = −⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦ ⎣ ⎦

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

The determinant of coefficient matrix is

19 0 100 7 310 3 20

A⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦

= 1789

By Cramer’s Rule

1

150 0 1050 7 3

0 3 20I

A

−⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥−⎣= ⎦ = -18150/1789 = -10145mA

By Cramer’s Rule

2

19 150 100 50 310 0 20

IA

−⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥⎣= ⎦ = -9500/1789 = -5.3102mA

By Cramer’s Rule

3

19 0 1500 7 5010 3 0

IA

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−⎣= ⎦ = 7650/1789=4.276mA

Current through R1 = I1 = -10.145mA

Current through R2 = I1 + I3 = -5.869mA

Current through R3 = I2 –I3 = -9.5862mA

Current through R4 = I2 = -5.3102mA

Current through R5 = I3 = 40276mA

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



Due Date: 24/12/2004

Q.1. Find Io in the network using Superposition. Draw and label the circuit diagram of each step, otherwise you will lose your marks. Write each step of calculation to get maximum marks.

Sol. Applying principle of superposition, we will take effect of the sources one by one Only current source is acting 2kΩ and 2kΩ are parallel so our new circuit will be

Now using current division rule I0I = (1)(9)/ 1+4 I0I = 1.8mA Only Voltage is acting

Now apply voltage rule VCD = (6) [1.33/(1.33+ 2)] VCD = 2.4 V I0V = 2.4/4kΩ I0V = 0.6 mA So, Io = I0I + I0V = 1.8 + 0.6 Io = 2.4mA

Q.2. Find Vo in the network using Source Transformation. Draw and label the circuit diagram of each step, otherwise you will lose your marks. Write each step of calculation to get maximum marks.

Sol.

-8 +I(4k+3k+3k)+9=0 I=-0.1mA V0 = (-0.1mA)(3k) = -0.3V

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



Due Date: 10/01/2005

Q.1. Find Vo in the network using Thevenin’s theorem. Draw and label the circuit diagram of each step, otherwise you will lose your marks. Write each step of calculation to get maximum marks.

Sol.

We want to calculate V0 by using Thevenin’s theorem .we will follow the following steps. First step: Removing RL

Here RL is 6k resistor across which we want to calculate the voltage Vo. Second step: Calculating VOC

We want to calculate VOC . Apply KVL in two loops to calculate the individual loop

currents. Here I2 = 2mA KVL for the Mesh 1 -6 +4kI1 +2k (I1 -2mA) =0

I1 = 10 /6K = 1.66mA

Vth =VOC = 4k I1 + 2KI2 = 10.66V Third step: Calculating Rth

Rth= 2k + (2k || 4k) = 3.33 K

Fourth step: Calculating the unknown quantity. After calculating Vth and Rth, re-inserting the load resistance RL in the circuit in

series with Rth and considering the Vth as a battery in series with these two

resistances.

V0 = 10.66 x 6k /( 6k +3.33k)

V0 = 6.85 volts

Q.2. Find Io in the network using Norton’s theorem. Draw and label the circuit diagram of each step, otherwise you will lose your marks. Write each step of calculation to get maximum marks.

Sol. We want to calculate I0 by using Norton’s theorem. We will follow these steps.

First step: Replacing RL with a short circuit to find ISC. Second Step:

For Node 1 V1= 12 V

For Node 2 (V2 – V1 )/6k + V2/2k + (V2 – 12) / 3 = 0

After solving we have V2= 6V Where ISC = (V1-V2)/3k +V1/4k = 5mA Third Step:

RN = Req = 4k ||[3k + (2k || 6k)] = 2.12K

Fourth Step:

After calculating IN and RN, re-inserting the load resistance RL in the

circuit in parallel RN and considering the IN or ISC current source parallel with these

two resistances.

IO=ISC [Req / (Req+2K)] = 2.57mA

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

Assignment 7 (Fall 2004)

Solution CIRCUIT THEORY (PHY301)

MARKS: 35

Due Date: 24/01/2005

Q.1.

The diagram below shows a simple diode circuit with a battery. In this particular case,

the diode is said to be forward-biased. Treat the diode as ideal (i.e. VB = 0). Sketch

the voltage drop over the resistor for

i. V = 0 ,

ii. V = ½Vs,

iii. V = Vs, and

iv. V = 2Vs.

The potential over the resistor is the sum of V(t) + VS minus the portion blocked by

the diode. Remember that if the potential is not greater than VBE, no current will get

through the diode. We will solve each case by sketching the potential before the diode

and then the sketch of the potential after the diode, that is everything in the first

graph that is greater than VBE.

i. V = 0

ii. V = ½V

iii. V = Vs

The diode is forward biased at all times, and there is no difference

between the curve above and the voltage across the resistor.

iv. V = 2Vs

The diode is forward biased at all times, and there is no difference

between the curve above and the voltage across the resistor.

Q.2.

A particular diode, for which n=1, is found to conduct 3mA with a junction

voltage of 0.7 V? to 0.8V? If the junction voltage is lowered to 0.69V ? to 0.6V? What

change in junction voltage will increase the diode current by a factor of 10.

Sol. Where VT = 0.025Vor 25mV

1. ID = 3mA = IS * exp( 0.7 / Vt ) = IS * exp( 0.7 / 25mV ) => IS = 2.08 x 10-15

A

Where 10-15

A = 1fA

2. ID = 2.08 fA * exp( 0.7 / 25mV ) = 3.008 mA

3. ID = 2.08 fA * exp( 0.8 / 25mV ) = 164 mA

4. ID = 2.08 fA * exp( 0.69 / 25mV ) = 2.01 mA

5. ID = 2.08 fA * exp( 0.6 / 25mV ) = 55uA

6.

n =1

VT = 0.025V

I1 =3mA

I2 = 30mA

As we know that

V = n VT ln(I2/ I1) --------(a)

Putting the values in equation (a)

V = (1) (0.025) ln(30/3)

= (0.025)ln(10)

= 0.025(2.305)

V = 0.057V

Q.3.

For the circuit in figure below using the constant-voltage-drop (VD = 0.7) diode

model, find the values of the labeled current and voltage when both diodes are on and

both off.

Assume all diodes off

1. No current flow

2. Diode drops are 10, 20

3. Diode 2 must be on

Both diodes are on

1. I(5k) = (10 - 0.7) / 5k = 1.86mA

2. I(10k) = (0.7 - 0.7 + 10) / 10k = 1.0mA

3. I(d1) = 1.86mA - 1.0mA = .86mA

4. V = 1mA * 10k - 10 = 0

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

Assignment 8 (Fall 2004) CIRCUIT THEORY (PHY301)

MARKS: 25

Due Date: 31/01/2005

Q.1.

The transistor in the circuit below has a very high . Find VE and VC for VB (a) +3V, (b)

+1V, and (c) 0V. Assume VBE 0.7 V.

Sol.

1. VB = 3V

1. VE = VB - 0.7 = 2.3V

2. IE = 2.3V / 1k = 2.3mA

3. VC = 9 - 2.3 * 1k = 6.7V

2. VB = 1V

1. VE = VB - 0.7 = 0.3V

2. IE = 0.3V / 1k = 0.3mA

3. VC = 9 - 0.3 * 1k = 8.7V

3. VB = 0.0 V

1. Base emitter can't forward bias, transistor cutoff, no current flow

2. VC = 9

3. VE = 0

Q.2.

Determine the dc load voltage and current values for the circuit shown in the fig.

Sol.

The transformer has 17 Vac

rating. Thus, the peak secondary voltage is found as

V2(pk)

= 17Vac

/0.707

V2(pk)

= 24.04 Vpk

The peak load voltage is now found as

VL(pk)

= V2(pk)

– 1.4

VL(pk)

= 22.64 Vpk

Vave

= 2Vpk

/ π

= 2(22.64) / π

Vave

=14.42VDC

Iave

= Vave

/ RL

=14.42/22

Iave =655micro A

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

Documents

Circuit Theory - Solved Assignments - Semester Fall 2004