Assignment 1 (Spring 2004) (Solution) CIRCUIT THEORY(PHY301) MARKS: 30 Due Date: 03/05/2004 Q.1. 1) Find the total resistance for the series circuit in the figure below 2) Calculate the source current I s . 3) Determine the voltages V 1 , V 2 ,V 3 and V 4 . 4) Calculate the power dissipated by R 1 , R 2 ,R 3 and R 4 . 5) 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 R T = R 1 +R 2 + R 3 + R 4 = 3 Ω + 2 Ω + 5 Ω + 7 Ω =17 Ω (b) Source Current I S = V/R T = 30/17 =1.76 A (c) Voltages will be V 1 = IR 1 = (1.76)(3 Ω) = 5.3V V 2 = IR 2 = (1.76)(2 Ω) = 3.52V V 3 = IR 3 = (1.76)(5 Ω) = 8.8V V 4 = IR 4 = (1.76)(7 Ω) = 12.32V (d) Power dissipated by each resistance P 1 = V 1 I 1 = (5.3) (1.76) = 9.33W P 2 = V 2 I 2 = (3.52) (1.76) = 6.2W P 3 = V 3 I 3 = (8.8) (1.76)
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Assignment 1 (Spring 2004) (Solution)
CIRCUIT THEORY(PHY301) MARKS: 30
Due Date: 03/05/2004
Q.1. 1) Find the total resistance for the series circuit in the figure below 2) Calculate the source current Is. 3) Determine the voltages V1, V2 ,V3 and V4. 4) Calculate the power dissipated by R1, R2 ,R3 and R4. 5) 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 Ω + 2 Ω + 5 Ω + 7 Ω =17 Ω (b) Source Current
IS = V/RT = 30/17 =1.76 A (c) Voltages will be
V1= IR1 = (1.76)(3 Ω) = 5.3V
V2= IR2 = (1.76)(2 Ω) = 3.52V
V3= IR3 = (1.76)(5 Ω) = 8.8V
V4= IR4 = (1.76)(7 Ω) = 12.32V
(d) 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)
= 15.5W P4= V4I4= (12.32) (1.76)
= 21.7W (e) 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
= (9Ω)(18Ω)/9Ω+18Ω = 162Ω/27
= 6Ω (b) IS= V/RT
= 27V/6Ω = 4.5A
(c) I1=V1/R1
= 27V/9Ω = 3A
I2=V2/R2
= 27V/18Ω = 1.5A IS = I1+I2
4.5A=3A+1.5A 4.5A=4.5A
(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= 6Ω x 6Ω/ 6Ω + 6Ω = 36/12 = 3Ω
RA = R1||R2||R3
= 3Ω x 2Ω/ 3Ω + 2Ω = 6/5
= 1.2Ω RB = R4||R5
= 8Ω x 12Ω/ 8Ω + 12Ω Fig A =9 6/20
= 4.8Ω Now the reduced form of the circuit will then appear as shown in figure B below
Fig B
RT = R1||R2||R3 + R4||R5
= 1.2Ω + 4.8Ω = 6Ω IS = V/RT
= 24V/6Ω = 4A V1= ISR1 || 2 || 3
= (4A)(1.2Ω) = 4.8V V5= ISR4 || 5
= (4A)(4.8Ω) = 19.2V Now Applying ohm’s law we have
I4 = V5/R4
= 19.2V/8Ω = 2.4A I2 = V2/R2 = V1/R2
= 4.8V/6Ω = 0.8A
------ Good Luck -----
Assignment 2 (Spring 2004) Solution
CIRCUIT 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.
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
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 = -0.74/9k
= V2/9k
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.
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 R1IR1 = I1 - I3 = 85.7mA - 28.57mA = 60.13mA
Current through R2IR2 = I1 - I2 = 85.7mA - 57.14mA = 28.56mA
Current through R3IR3 = I2 - I3 = 57.14mA - 28.57mA = 28.57mA
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 ) (150Ω) = 6.3448 V ER2=IR2 R2 = (93.793 mA)(50Ω) = 4.6897 V ER3=IR3 R3 = (16.552 mA)(100Ω) = 1.6552V ER4=IR4 R4 = (58.851 mA)(300Ω) = 17.6552V ER5 =IR5 R5 = (77.241 mA)(250Ω) = 19.3103V
Q.3. Solve for the mesh currents in the circuit below.
Q.4.Find Ia and Ib in the following circuit using mesh analysis. Identify and label each mesh.
Sol.
First, label the mesh currents
Then, notice that there are two current sources and therefore two of the mesh currents can be eliminated.
I4 = 5 A
I3 = -2.5 I a + I2 = -2.5 I2 + I2 = -1.5 I2
There are also two dependant sources which can be simplified.
2 I b = 2 (I3 - I1) = 2 (-1.5 I2 - I1)
2 I b = -2 I1 - 3 I2
2.5 I a = 2.5 I2
For Mesh 1:
Then, find the equations which describe the two currents. First, i1 opposes i2 in a 40 ohm resistor, opposes i3 in a 8 ohm resistor, and rises 2 ib V due to a voltage source.
40 (I1 - I2) + 8 (I1 - I3) = 2 I b
40 (I1 - I2) + 8 (I1 + 1.5 I2) = -2 I1 - 3 I2
40 I1 - 40 I2 + 8 I1 + 12 I2 = -2 I1 - 3 I2
50 I1 - 25 I2 = 0
2 I1 - I2 = 0 ----------- (A)
For Super Mesh :
Next, find look at the Super mesh formed by combing the meshes of I 2 and I3. First, I2 opposes I1 in a 40 ohm resistor, I3opposes I4 in a 80 ohm resistor, I3 opposes I1 in a 8 ohm resistor, and there is a 64 V rise due to a voltage source.
2 1 A sc Ni i I∴ = = = After calculating IN and RN, the Norton equivalent circuit is,
Q.2. Using the superposition theorem, find io in the following circuit.
Sol.
' ''
o o oi i i= +
20V 4A
(i) When 4A source active only → voltage source shorted , we suppose that current = i’
o
From Loop 1:
1 4 A............(1)i =
From Loop 2:
'2 1 2 2 33( ) 2 1 1 5 0oi i i i i i− + + − − =
'1 2 33 6 1 5 0.....(2)oi i i i− + − − =
From Loop 3:
'
3 1 3 2 35( ) 1( ) 4 5 0oi i i i i i− + − + + =
'
1 2 35 1 10 5 0.......(3)oi i i i− − + + =
From the Figure it is clear that
At Node 0:
' '
3 1 4 ..........(4)o oi i i i= − = − (1) & (4) → (2) & (3)
'2
'2
3 25 2
o
o
i ii i⎧ − =⎨
+ =⎩
80
' 52 A17oi∴ =
(ii) When 20V Source active only→ current source opened, we suppose that current = i’’
o
From Loop 4:
''
4 56 5 0.......(5)oi i i− − =
From Loop 5:
''
4 510 20 5 0....(6)oi i i− + − + =
''
5But (5) & (6)oi i= − →
''4
''4
6 4 05 2
o
o
i ii i⎧ − =⎨
+ = −⎩ 0
'' 60 A
17oi∴ = −
' '' 8 0 .4 7 0 6 A
1 7o o oi i i= + = − = −
Q.3. Use source transformation to find vx in the following circuit.
Sol.
We can solve above circuit by two methods: Method: I From large loop 3 (1 4) 18 0.......(1)xi v− + + + + =From small loop 3 1 0 3 .......(2)x xi v v i− + + = ⇒ = − Put value from 2 in (1) we have 15 5 3 0 4.5 Ai i i+ + − = ⇒ = −Alternatively Method: II
From right small loop 4 18 0 4.5 Ax xv i v i− + + + = ⇒ = −
Put i= -4.5 A in (2) we have 3 3 ( 4.5) 7.5 Vxv i∴ = − = − − =
Q.4. Find the Thevenin equivalent circuit of the following circuit, to the left of the terminals a-b.
1 2 1 2
2 1 2
1 2
2 2( ) 0 But 4 3 ..............(1)
x x
x
v i i v ii v i i
i i
− + − = ⇒ = −− = = −
∴ = −
i
2 2 1 2 34 2( ) 6( ) 0........(2)i i i i i+ − + − =
3 2 36( ) 2 1 0.................(3)i i i− + + =
Sol. (A) Find RTh
Loop 1:
Loop 2:
Loop 3: Solve (1)(2) and (3) we have
31 A6
i = −
3 Th0
1 1 VBut A, 66oi i R
i= − = ∴ = = Ω
(B) Find VTh
By mesh analysis Loop 1:
1 5 A............(1)i = Loop 2:
2 1 2 3 2
2 1 3
4( ) 2( ) 6 0 12 4 2 0.............(2)
i i i i ii i i
− + − + =⇒ − − =
Loop 3: 3 2
3 2
2 2( ) 0 .............(3)x
x
v i iv i i
− + − =⇒ = −
1 2But 4( )...............(4)xv i i= −
Solving (1) , (2) ,(3) and (4) simultaneously we have
210 A3
i∴ =
Th 26 20 VocV v i= = =
Thevenin’s equivalent circuit will be
------ Good Luck -----
Assignment 6 (Spring 2004) (Solution)
CIRCUIT THEORY (PHY301) MARKS: 30
Due Date: 24/07/2004 Note: Labeled and draw each circuit diagram if necessary, other wise you will lose your marks. Show how you got your equations. Be sure to show how you got your answers.
Q.1. For the following circuit, assume io = 1 A and use LINEARITY PROPERTY to find the actual value of io .
=? A
Sol. We will redraw the figure
is = 3 A ⇐ io = 1 A
is = 15 A ⇒ io = 5 A If io = 1A, then v1 = (3+5) io = 8 V i1 = v1 / 4 = 2A
KCL at Node 1:
D10 - 0.7I = = 0.93 mA
10
KCL at Node 2: 4 3 2 45A i 5 Asi i i i= + = ⇒ = =
io = 1 A → is = 5 A io = 3A ← is = 15 A
Q.2.
(a) Find voltage across R1, consider germanium diode so that 0.3 drop is assumed across it.
Sol. By KVL
Vs=VR+Vf VR=Vs- 0.3
VR= 9 - 0.3
= 8.7 volts .
(b) Determine the voltage VF across the diode in the figure below for values of IF = 3mA and IF =7 mA. Assume that the bulk resistance of the diode is 7 Ω and the diode is silicon. .
Sol. Because the diode is silicon we use VF = 0.7 V+ IF RB For IF = 3mA VF = 0.7 + (3mA) (7Ω) = 0.7 + 0.021V = 0.721 V (721 mV) For IF =7mA VF = 0.7 + (7mA) (7Ω) = 0.7 + 0.049 V = 0.749 V (749 mV)
Q.3. Consider the circuit shown in the figure below for the case R=10k Ω. The power supply V+ has a dc value of 10 V on which is super imposed a 60-Hz sinusoid of I-V peak amplitude. (This “signal” component of the power supply voltage is an imperfection in the power supply design. It is known as the power supply ripple.) Calculate the dc voltage of the diode. Assume the diode to have 0.7-V drop at 1mA current and n =2
Sol. Considering dc quantities only, we assume VD ≅ 0.7V and calculate the diode dc current
D10 - 0.7I = = 0.93 mA
10
Since this value is very close to 1mA, the diode voltage will be very close to the assumed value of 0.7 V. At this operating point , the diode incremental resistance rd is
T
dD
nV 2 25r = = = 53.8I 0.93
×Ω
The peak signal voltage across the diode can be found by using the voltage divider rule
dd
d
rV (peak to peak) = 2 R + r
0.0538= 2 = 10.7 mV10 +0.0538
------ Good Luck -----
Assignment 7 (Spring 2004) (Solution)
CIRCUIT THEORY (PHY301) MARKS: 25
Due Date: 01/08/2004
Q.1. With the sine wave signal input of Fig (a), find the shape of the output signal VO in the biased series clipper of Fig (b). What would happen if battery connections were reversed? Also draw the shape of out signal.
Sol. Let us consider the positive half- cycle of the signal i.e. , when terminal A of the circuit becomes positive with respect to B. The diode appears as a short since it is forward-biased but no current flows till Vg exceeds the opposing battery of 10 V. Hence only upper part of the positive signal voltage passes through the clipper circuit and appears as vo across R. Its shape is shown in the figure below. The negative half cycle of the signal voltage is clipped off. In fact, in this circuit, the entire input is clipped off except positive peak portions.
Reverse Battery Connections The battery connections have been reversed in the Following figure. In this case, during the positive half-cycle of the signal, the voltage across R would be the sum of the signal voltage and the battery voltage i.e. signal voltage would be lifted up by 10V as shown in the fig(c). During the negative input half-cycle , the lower peak portion of the signal would be clipped off because of the battery. Q.2. Determine the peak and dc load current values for the circuit below.
Sol. The transformer is rated at 40Vac. Therefore, the value of V2(pk) is found as V2(pk) = 40Vac/0.707 = 56.6Vpk
The peak load voltage can now found as VL(pk) = V2(pk) /2 – 0.7 =28.3 – 0.7 = 27.6Vpk
Finally the dc load voltage is found as Vave = 2 VL(pk) /π
= 55.2 Vpk/ π Vave =17.57Vdc IL(pk) = VL(pk)/RL = 27.6 / 8.3 k = 3.33 mApk And Iave = Vave/RL =17.57/8.3 k = 2.12 mA dc
Q.3. A full wave rectifier has 16 Vac transformer and 16 k Ω load. Determine the dc load voltage and current values for the circuit.
Sol.
The peak secondary voltage is found as V2(pk) = 16/0.707 =22.637Vpk
The peak load voltage is now found as VL(pk) = V2(pk) /2 – 0.7V VL(pk) =10.62Vpk The dc load voltage is found as Vave = 2VL(pk)/π
= 21.24/ π Vave =6.76 Vdc Finally the dc load current is found as Iave = Vave/RL = 6.76 /16 k Iave = 422.5 microA
------ Good Luck -----
Assignment 8 (Spring 2004) (Solution)
CIRCUIT THEORY (PHY301) MARKS: 10
Due Date: 08/08/2004 Q.1. Calculate the currents and voltages in a circuit when the transistor is driven into
saturation. For the circuit shown in Figure below the transistor parameter are β=100, and VBE(on) =0.7 V. If the transistor is biased in saturation, assume
VCE (sat) =0.2 V.
Sol.
Since +8V is applied to the input side of RB, the base-emitter junction is certainly forward biased , so the transistor is turned on. The base current is
BB BEB
B
V -V 9- 0.7I = = 33.2 AR 220
µ⇒
If we first assume that the transistor is biased in the active region, then the collector current is
C BI I (100)(33.2 A) 3.32 Amβ µ= = ⇒
The collector- emitter voltage is then
CE CC C CV V I R 10 (3.3)(4) 3.28V= − = − = −
However the collector emitter voltage of the npn transistor in the common-emitter configuration shown in the above figure cannot be negative. Therefore, our initial assumption of the transistor being biased in the forward- active mode is incorrect. Instead, the transistor must be biased in saturation.
As given in the question statement , set VCE(sat) =0.2V. The collector current is
CC CEC C
C
V -V (sat) 10-0.2I = I (sat) = = 2.45mAR 4
=
Assuming that the B-E voltage is still equal to VBE(on)=0.7V, the base current is IB=33.2µA, as previously determined. If we take the ration of collector current to base current, then