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BITS PILANI, K.K. BIRLA GOA CAMPUS ELECTRICAL SCIENCES (EEE F111)
Tutorial -2 on 28-01-2015
1) Find value of current i1 in the following circuit using
superposition theorem.
ANS: With only 30V voltage source active
Using mesh analysis: Apply KVL to each mesh,
-30+6I1+VX=0 & -Vx+4 I2+2 I2=0
Vx=30-6I1 & Vx= 6I2
Since LHS are equal,RHS of two mesh equations are equal
30-6I1=6I2.-----------equn 1
Also, I1-I2=8i11 & I2=i11-----Substituting in above equn-1
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30=60 i11 Hence i11=0.5A
With only 3A.current source active
6I1+Vy=0 & -Vy+4 (I2-I3)+2 I2=0
Vy=-6I1 & Vy= 6I2 - 4I3
Since LHS are equal, RHS of two mesh equations are equal
6I1=6I2 - 4I3.-----------equn 2
Also, I1-I2=8i12 & I2=i12 & I3=3 A---Substituting in equn-2
12=60 i12 Hence i12=0.2A
Using Superposition theorem, i1=i11+i12=0.5+0.2=0.7A
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Resistors 30 ohms in parallel with 15 ohms.(30 || 15 = 10 ohms)
KVL equation around mesh is : 10I+150+10I+[(1/3)vx3]=0 & since Vx3=(-10I).
Solving these two equations, 20I+150-10I=0; 60I+450-10I=0; 50I+450=0; I=-9A;
Vx3=-10I=90V. Hence Vx=15-30+90=75V-------Ans.
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Q. In the circuit shown, find the value of R that will absorb maximum power from the circuit. Find also the value of this maximum power.
Solution: ThR R= :
KVL:
3 8( 1) 0 5
4 A 40 V 40 1 A
x x x
tt Th
V V i V iVi V R
+ + + = =
= = = =
VTh:
KVL: 2 21
3 8 04x x x
V V i i V + + = =
1 2 15
5( ) 2 A 104
40 V, 2 80 V
x x x
x Th x
V i i i V V
V V V
= = = +
= = =
2
max
640040 W
4 4 40ThVPR
= = =
+
Vx
5 R8 2 A
+ 3Vx
+
Vx
5 8
+ 3Vx
1 A
+
Vt
i
+
Vx
5 8 2 A
+ 3Vx
+
VTh
i1 i2
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Q. The variable resistor oR in the circuit in Fig.3 is adjusted until it absorbs maximum power from the circuit.
o
2
th
a) Find the value of R .
b) Find the maximum power.
10 V3 A2 A Ro
4 : 6 :
Ro
Rth
Vth
When o thR R it will absorb maximum power. Figure 3
Maximum power is max4o
thR
VP
R
4 : 6 :
Rth = 10 :
In order to find thR all independent
sources are killed.
For V : th
KCL at V :th
24
8.............(1)
th
th
V V
V V
10 V3 A2 A
4 : 6 : 10 VVth V
KCL at V :
103
4 6
3 3 2 20 36
3 5 16..........(2)
th
th
th
V V V
V V V
V V
Multiply Eq. (1) by 5 and then sum up Eq(1) and (2)
max
2 24
12 V
10
1443.6
40o
th
th
o
R
V
V
R
P W
:
Tut-2Tut-2tut2
max_power2max_power3