Problem 3.5
Solution:
Known quantities:
Circuit shown in Figure P3.5 with mesh currents: I1 = 5 A, I2 = 3 A, I3 = 7 A.
Find:
The branch currents through:
a)
R1,
b)
R2,
c)
R3.
Analysis:
a) Assume a direction for the current through R1 (e.g., from
node A to node B).
I R1 = I1 − I 3 ( I 3 has the opposite direction to I R1 )
I R1 = I1 − I 3 = −2 A
“ – ” means that the assumed direction of the current through R1 is not correct.
b) Assume a direction for the current through R2 (e.g., from node B to node A).
I R2 = I 3 − I 2 = 4 A
The assumed direction of the current through R2 and the direction of I3 are the same.
c) Only one mesh current flows through R3. If the current through R3 is assumed to flow in the same direction,
then:
I R1 = I 3 = 7 A .
Problem 3.11
Solution:
Known quantities:
The voltage source value, 3 V, and the five resistance values, indicated in Figure P3.11.
Find:
The current, i, drawn from the independent voltage source using node voltage analysis.
Analysis:
n=4, m=1, Total number of variables: n-1-m = 2: two nodes and two equations for node 1 and 2
At node 1:
i = i A + iB →
3 − v1 v1 v1 − v 2
=
+
→ 3 = 4v1 − 2v 2
0.5
0.5
0.25
0.75 Ω
v = 3V
iB
At node 2:
i B = 0. 5 + i C →
iC
v1 − v2
v
= 2 + 0.5 → 3v1 − 4v2 = 0.5 × 0.75
0.25
0.75
Solving the system, we obtain:
v1 = 1.125 V , v 2 = 0.75 V Therefore, i =
iA
v=0V
3 − v1
= 3.75 A .
0.5
Problem 3.12
Solution:
Known quantities:
Circuit shown in Figure P3.12.
Find:
Power delivered to the load resistance.
Analysis:
n=4, m=1, Total number of variables: n-1-m = 2: two nodes and two equations for node 1 and 2
KCL at node 1:
V -(V + V )
V
I S = i A + i B → 1 + 1 s 2 = 0.5
RI
RV
Or 3 V1 - V2 = 6
14 V2- 4 V1 = -16
substitute Eq. 1 into Eq. 2
and by voltage divider:
 RL 
V2 = −0.316V
VL = 
 R2 + RL 
iD
iB
KCL at node 2:
V -(V + V ) V
V2
i B = iC + i D → 1 s 2 = 2 +
RV
R1 (R2+R L )
Or
( Eq. 1)
iC
iA
( Eq. 2)
ground
V2 = − 12 / 19 Volt = -0.6316Volt
PL =
VL2
= 25 mW
RL
Problem 3.20
Solution:
Circuit shown in Figure P3.20.
Find:
Mesh equation in matrix form and solve for currents.
Analysis:
n=4, m=1, Total number of variables: n-m = 3.
−4
− 4  i2  9
KVL for i2 : 3 + 6(i2 − 2 ) + 4(i 2 − i 4 ) + 4(i2 − i3 ) = 0 6 + 4 + 4

 
   
4+ 4+8
0  i3  = 3
KVL for i3 : −3 + 4(i3 − i2 ) + 4i3 + 8i3 = 0
→  −4
KVL for i : −5 + 2i + 4(i − i ) = 0
  −4
0
2 + 4 i4  5
4
4
4
2

 
i1 = 2 A
Problem 3.27
Solution:
Known quantities:
The values of the resistors and of the
voltage source in the circuit of Figure
P3.27.
Find:
The current i through the resistance
R 4 mesh current analysis.
Analysis:
n=3, m=1, Total number of variables: n-m = 2.
For mesh (a): KVL
− 5.6 + 50ia + 1200(ia − ib ) = 0 → ia (50 + 1200) + ib (− 1200) = 5.6
For meshes (b) and (c): KVL
330ib + 1200(ib − ia ) + v(= 440ic ) = 0 → ia (− 1200) + ib (1200 + 330) + ic (440) = 0
For the current source: KVL -> constitutive equation
i c − i b = 0.2v x = 0.2 (1200 (i a − i b )) = 240 (i a − i b )
Solving,
i a = 136 mA , i b = 137 mA and i c = −106 mA .
Therefore,
i = i c = −106 mA .
v
Problem 3.41
Solution:
Known quantities:
The values of the current source, of the voltage source and of
the resistors in the circuit of Figure P3.41:
I B = 12A RB = 1Ω VG = 12 V RG = 0.3Ω R = 0.23Ω
+
Find:
The voltage across R1 using superposition.
Analysis:
Specify a ground node and the polarity of the voltage across R. Suppress the voltage source by replacing it with
a short circuit. Redraw the circuit.
1
1
IB
R
R
By current dividers: I R =
I B → VR − I =
IBR =
= 1.38 V
1
1
1
1
1
1
1
1
1
+
+
+
+
+
+
R B RG R
RB RG R
RB RG R
Suppress the current source by replacing it with an open circuit.
(RB || R ) V = (0.187) 12 = 4.61 V
By voltage dividers: VR −V =
G
RG + (RB || R )
0.3 + (0.187 )
By superposit ion : V R = VR − I + VR −V = 5.99 V
Problem 3.55
Solution:
Known quantities:
Circuit shown in Figure P3.55.
Find:
Thevenin equivalent circuit
Analysis:
To find RT, we zero the two sources by shorting the voltage source and opening the current source. The resulting
1k Ω
circuit is shown in the left:
1Ω
3Ω
a
Therefore, RT = 1,000|| 1,000 + 1 + 3 = 504 Ω .
To find Thevenin voltage Voc , we assume Vb as ground (i.e.,
1k Ω
zero) and apply nodal analysis.
Vc − Va
= 0.01 → Va = −0.01 V
1
10 − Vc
V
= c + 0.01 → 10 − Vc = Vc + 10 → Vc = 0 V
1000
1000
Therefore, VOC = −0.01 V .
It means that a is negative side, b is positive side while the
magnitude of Voc is 0.01V.
Problem 3.59
Solution:
Known quantities:
Circuit shown in Figure P3.59.
Find:
Value of resistance Rx
Analysis:
Vab =
Rx
R
VS VS
R+R
R+Rx
Rx
1
VS VS
2
R + Rx
b) For R = 1 kW , Vs = 12 V , Vab = 12 mV ,
Vc
1Ω
3Ω
0V
10V
Then, Va = −0.01V
a) We have Vab = Va -Vb =
b
1kΩ
1kΩ
10mA
+
Va
Voc
-
Vb
0.012 = 6-
Rx
12
1000+Rx
Rx = 996Ω
Problem 3.73
Solution:
Known quantities:
The values of the voltage and of the resistor in the equivalent
circuit of Figure P3.73: VT = 12V; RT = 8Ω
Assumptions:
Assume the conditions for maximum power transfer exist.
Find:
a.
b.
c.
The value of RL .
The power developed in RL .
The efficiency of the circuit, that is the ratio of power absorbed by the load to power supplied by the source.
Analysis:
a.
For maximum power transfer: RL = RT = 8Ω
b.
VD:
VRL =
RL
(8) (12) = 6V
VT =
RT + R L
8+8
V 2 RL (6)
=
= 4.5 W
RL
8
2
PRL =
c.
η=
P0
I 2R
PRL
RL
=
= 2 S L2
=
= 0.5 = 50%
PS PRT + PRL I S RT + I S RL RT + R L