Circuit Theory - Solved Assignments - Spring 2004

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Circuit Theory - Solved Assignments - Spring 2004

Text of Circuit Theory - Solved Assignments - Spring 2004

Assignment 1 (Spring 2004)(Solution)

CIRCUIT THEORY(PHY301) MARKS: 30 Due Date: 03/05/2004

Q.1.1) 2) 3) 4) 5) Find the total resistance for the series circuit in the figure below Calculate the source current Is. Determine the voltages V1, V2 ,V3 and V4. Calculate the power dissipated by R1, R2 ,R3 and R4. Determine the power delivered by the source, and compare it to the sum of the power levels of part (4)

Sol.(a) Total resistance will be RT = R1 +R2 + R3 + R4 = 3 + 2 + 5 + 7 =17 Source Current IS = V/RT = 30/17 =1.76 A

(b)

(c)

Voltages will be V1= IR1 = (1.76)(3 ) = 5.3V V2= IR2 = (1.76)(2 ) = 3.52V V3= IR3 = (1.76)(5 ) = 8.8V V4= IR4 = (1.76)(7 ) = 12.32V Power dissipated by each resistance P1= V1I1= (5.3) (1.76) = 9.33W P2 = V2I2= (3.52) (1.76) = 6.2W P3 = V3I3= (8.8) (1.76)

(d)

(e)

= 15.5W P4= V4I4= (12.32) (1.76) = 21.7W Pdel = EI =VI =(30)(1.7)= 53 W Pdel = P1 + P2+ P3 + P4 53 = 53 (checked)

Q.2.For the parallel network of figure below 1) Calculate the RT. 2) Determine Is. 3) Calculate I1 and I2 and demonstrate that Is=I1 + I2. 4) Determine the power to each resistive load. 5) Determine the power delivered by the source, and compare it to the total power dissipated by the resistive elements.

Sol.(a) Total resistance will be RT = R1R2 / R1 + R2 = (9)(18)/9+18 = 162/27 = 6 IS= V/RT = 27V/6 = 4.5A I1=V1/R1 = 27V/9 = 3A I2=V2/R2 = 27V/18 = 1.5A IS = I1+I2 4.5A=3A+1.5A 4.5A=4.5A

(b)

(c)

(d)

P1= V1I1 = (27V) (3A) = 81W P2 = V2I2= (27) (1.5A) = 40.5W

(e)

Pdel = EIS =VI =(27)(4.5)=121.5W Pdel = P1 + P2 121.5W = 81W + 40.5W =121.5W

Q.3.Find the indicated currents and voltages for the network given below.

Sol.R1||R2= 6 x 6/ 6 + 6 = 36/12 = 3 RA = R1||R2||R3 = 3 x 2/ 3 + 2 = 6/5 = 1.2 RB = R4||R5 = 8 x 12/ 8 + 12 Fig A =9 6/20 = 4.8 Now the reduced form of the circuit will then appear as shown in figure B below

Fig B RT = R1||R2||R3 + R4||R5 = 1.2 + 4.8 = 6 IS = V/RT = 24V/6 = 4A V1= ISR1 || 2 || 3 = (4A)(1.2) = 4.8V V5= ISR4 || 5 = (4A)(4.8) = 19.2V Now Applying ohms law we have

I4 = V5/R4 = 19.2V/8 = 2.4A I2 = V2/R2 = V1/R2 = 4.8V/6 = 0.8A

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

Assignment 2 (Spring 2004)SolutionCIRCUIT THEORY(PHY301) MARKS: 30 Due Date: 11/05/2004

Q.1.Find VAF and VCH in the circuits from all possible paths.

Sol. We will draw some imaginary arrows form point A to point F and form point C to H.

VCH

VAF

For Path AFGHA VAF -10 + 1 12 = 0 --------------------- (A) VAF = 21Volts For Path ABCDEFA 2 +8+5+4+2+ VFA= 0 -------------------- (B) VAF= -VFA -VFA = 21Volts VAF = 21Volts To calculate VCH, we will write equations for Path CHABC and CDEFGHC For Path CHABC VCH 12 + 2 + 8 = 0 --------------------- (C) VCH = 2Volts

For Path CDEFGHC 5 + 4 + 2 10 + 1+ VHC = 0 ------------------ (D) VCH= -VHC Hence, VCH =2Volts

Q.2.(a) Use nodal analysis to find the current I0 through 6k ohm resistance.

Sol.12k

V1

6k

V2

+ 8V -

7k

12k

3mA

At node 1 (V1-8)/12 + V1/7 + (V1 V2)/6 = 0 (7V1 -56 + 12V1 +14V1 14V2)/84 = 0 (7V1 -56 + 12V1 +14V1 14V2) = 0 33V1 14V2 = 56 ---------------- (A) At node 2 (V2 V1)/6 + V2/12 = -3mA 2V2 2V1 +V2 = -36 -2V1 +3V2 =-36 ----------------- (B) Multiply equation (A) with 2 and (B) with 33 and then adding we have 66V1 - 28V2 = 112 - 66V1 + 99V2 = -1188 71V2 = -1076

V2= - 15.15 V Put the value of V2 in B we have -2V1 +3(-15.15) = - 36 -2V1= 45.45-36 V1= -4.7 V Now we know the voltage at both nodes so now we will calculate the value of current through 6K resistance. So for output current Io I = (V1 V2)/6o

= (-4.7 +15.15)/6 = 10.45/6 =1.74mA

(b) Use nodal analysis to find the current I0 through 6k ohm resistance.

Sol.16k

V1

6k

+ 14V 8k

+ 8V -

At node 1 (V1-(-14)/16k + V1/8k + (V1-8) /6k = 0 3V1+ 42 +6V1 + 8V1 -64 = 0 17V1 = 22 V1= 1.294V Io = (8-V1))/6k Io= (8-1.294)/6k = 1.12 Io=1.12mA

Q.3.Use nodal analysis to calculate the voltage V0 and current I0.

Sol.

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

Node 2 and 3 constitute a super node Constraint equation will be V3 V2 = 8 -------------------- (A) KCL equation at super node is ((V2-10)/6k) + (V2/9k) + (V3 /3k) + ((V3 -10)/5k)=0 After simplifying we have 25V2 +48 V3=330 ----------------------------(B) After solving equation (A) and (B), we have V3 = 7.26Volts V2 = -0.74Volts V0 = V3 = 7.26Volts

I0 = V2/9k = -0.74/9k I0 = - 0.08222mA

Assignment 3 (Spring 2004) (Solution)CIRCUIT THEORY (PHY301) MARKS: 30 Due Date: 19/05/2004

Q.1.Use Mesh analysis to find currents through all loops also find currents through each resistance in the networks below.

(a)

Sol.

For Loop 1 -20 + 100 (I1 - I3) + 500 (I1 - I2) = 0 600*I1 - 500*I2 - 100*I3 = 20 ---------------- (A) For Loop 2 500 (I2 - I1) + 100 (I2 - I3) + 200 (I2) = 0 -500*I1 + 800*I2 -100*I3 = 0 --------------- (B) For Loop 3 300 (I3) + 100 (I3 - I2) + 100 (I3 - I1) = 0 -100*I1 - 100*I2 + 500*I3 = 0 --------------- (C)

Subtracting (A ) and (B) we have 1100 I1-1300I2=20 ------ (1) Multiply equation (A) by 5 and then adding in (C) we have 2900 I1 2600 I2 =100 ----- (2) Multiply equation (1) by 2 and subtracting by (2) we have I1 = 60 / 700 = 85.7 mA Put this value in (1) we have I2= 57.14 mA After putting the value of I1 and I2 in (A) we have I3 = 28.57mA Current through R1 IR1 = I1 - I3 = 85.7mA - 28.57mA = 60.13mA Current through R2 IR2 = I1 - I2 = 85.7mA - 57.14mA = 28.56mA Current through R3 IR3 = I2 - I3 = 57.14mA - 28.57mA = 28.57mA Current through R4 IR4 = I3 = 28.57mA Current through R5 IR5 = I2 = 57.14mA

(b)

Sol.

For Loop 1: -58 + 10*I1 + 6 (I1 - I2) = 0 16*I1 - 6*I2 = 58 --------------- (A) For Loop 2: 6 (I2 - I1) - 6 + 4*I2 + 5 (I2 + I3) =0 -6*I1 + 15*I2 = -9 ------------- (B) For Loop 3: I3 = 3A Solving (A) and (B) for I1 and I2: I1 = 4A I2 = 1A IR1 = I1 = 4A IR2 = I1 - I2 = 4 - 1 = 3A IR3 = I2 = 1A IR4 = I2 + I3 = 1 + 3 = 4A

Q.2.Use Mesh analysis to find currents and also calculate voltage drop across each resistor.

Sol.

For Loop 1: -50I1 - 100(I1 +I2) - 150(I1 + I3) =0 300I1 + 100I2 + 150 I3 =0 ----- (A) For Loop 2: -100(I1 +I2) - 300(I2 I3) - 250 I2 = 0 100I1 + 650I2 -300 I3 =0 ----- (B) For Loop 3: 24 -150(I3 +I1) - 300 (I3 I2) = 0 -150I1 + 300I2 -450 I3 = -24 ----- (C) Simultaneously solve (A) (B) and (C) I1 = - 93.793 mA I2 = 77.241 mA I3 = 136.072 mA The negative value arrived at for I1 tells us that the assumed direction for that mesh current was incorrect. Thus, the actual current value through each resistor is as such:

IR1 = I3 - I1 = 136.092 mA 93.793 mA4A = 42.299 mA IR2 = I1 = 93.793 mA IR3 = I1-I2 = 93.793 mA - 77.24 mA = 16.552 mA IR4 = I3 - I2 = 136.092 mA - 77.241 mA = 58.851mA IR5 = I2 = 77.241mA Calculating voltage drops across each resistor: ER1=IR1 R1 = (42.299 mA ) (150) = 6.3448 V ER2=IR2 R2 = (93.793 mA)(50) = 4.6897 V ER3=IR3 R3 = (16.552 mA)(100) = 1.6552V ER4=IR4 R4 = (58.851 mA)(300) = 17.6552V ER5 =IR5 R5 = (77.241 mA)(250) = 19.3103V

Q.3.Solve for the mesh currents in the circuit below.

Sol.

I3

I1

I2

Mesh 1:

3I1 + 10(I1 I3) + 4(I1 I2) = 20 + 10 17I1 4I2 10I3 = 30 ---------------- (A) 4(I2 I1) + 11(I2 I3) + 3I2 = - 10 - 8 -4I1 + 18I2 11I3 = -18 --------------- (B)

Mesh 2:

Mesh 3:

9I3 + 11(I3 I2) + 10(I3 I1) = 12 + 8 -10I1 11I2 + 30I3 = 20 ---------------- (C)

Multiply (A) by 9 and (B) by 2 and adding (A &B) we have 153I1 36I2 90I3 = 270 -8I1 + 36I2 22I3 = -36 145I1 112I3 = 234 I1= (234 + 112 I3)/ 145 ----------- (1)

Multiply (A) by 4 and (B) by 17 and adding we have 68I1 16I2 40I3 = 120 -68I1 + 306I2 187I3 = -306 290I2 - 227I3 = -186 I2 = (-186 +227 I3)/290 ---------------(2) After putting the values of I1 and I2 in (C ) we have

I3 = 2.13A Put I3 in (1) and (2) we have I1 = 3.25 A I2 = 1.03 A ------ Good Luck -----

Assignment 4 (Spring 2004) (Solution)CIRCUIT THEORY (PHY301) MARKS: 35 Due Date: 24/06/2004

Q.1. Use mesh analysis to find the current io. Identify and label each mesh. Solve equations by using matrices.

Sol. First, label the mesh currents.

KVL for Mesh 1: -24 +10 (i1- i2) + 12(i1- i3) = 0 11i1 5i2 6i3 = 12 ----------- (1) KVL for Mesh 2: 24 i2 +4 (i2- i3) + 10 (i2- i1) = 0 -5i1 + 19i2 2i3 = 0 ----------- (2)

KVL for Mesh 3: 4 i0 + 12 (i3- i1) + 4(i3- i2) = 0 But i0= i1- i2 4(i1- i2) + 12 (i3- i1) + 4(i3- i2) = 0 - i1 i2 +2 i3 = 0 ----------- (3)

In matrix form 11 5 6 i1 12 5 19 2 i = 0 2 1 1 2 i3 0

11 5 6 D = 5 19 2 1 1 2 = 11{(38-2)} (-5){(-10 -2)} + (-6){(5 +19)} = 192 12 5 6 D1 = 0 19 2

0 1

2

12 5 6 0 19 2 I1 = 1 2 192 = 12 (38 -2)/192 = 2.25 A 11 12 6 0

D2 = 5 1

0 2 0 2

11 12 6 5 0 2 I2 = 0 2 192 = -12 (-10 - 2)/192 = 0.75 A 11 5 12 1

D3 = 5 19 1 1

0 0

11 5 12 5 19 0 1 1 0 192 = 12 (5 + 19)/192 = 1.5 A io = i1 i2 = 1.5 A Q.2. Identify and label each mesh. Find current through all meshes using mesh analysis for the following circuit. I3 =

Sol.

First, label the mesh currents.

KVL to the larger Super mesh: 2i1 + 4i3 + 8(i3-i4) + 6i2 = 0 i 1 + 3i2 + 6i3 - 4 i4 = 0 -------------- (1) Constraint equation for for independent source i2 = i 1 + 5 --------------- (2) Constraint equation for dependent source i2 = i 3 + 3 i 0

but i o = - i 4

i2 = i 3 - 3i4 --------------- (3) KVL to mesh 4: 2i4 + 8(i4-i3) + 10 = 0 5i 4 - 4 i3 = -5 -------------- (4) Solve (1) to (4) i 1 = - 7.5 A, i 2 = - 2.5 A, i 3 = 3.93 A, i 4 = 2.143 A

Q.3. Find I in the