P2.68
Writing KVL equations around each mesh, we have
5i1 + 7(i1 − i3 ) + 31 = 0
11(i2 − i3 ) + 3i2 − 31 = 0
i3 + 11(i3 − i2 ) + 7(i3 − i1 ) = 0
Putting the equations into standard from, we have
12i1 − 7i3 = −31
14i2 − 11i3 = 31
− 7i1 − 11i2 + 19i3 = 0
Using Matlab to solve, we have
>> R = [12 0 -7; 0 14 -11; -7 -11 19];
>> V = [-31; 31; 0];
>> I = R\V
I=
-2.0000
3.0000
1.0000
Then, the power delivered by the source is P = −31(i1 − i2 ) = 155 W.
P2.75
(a) First, we select mesh-current variables as shown.
Then, we can write
(Rw + Rn + R1 )i1 − Rn i2 − R1i3 = 120
− Rn i1 + (Rw + Rn + R2 )i2 − R2i3 = 120
− R1i1 − R2i2 + (R1 + R2 + R3 )i3 = 0
Alternatively, because the network consists of independent voltage
sources and resistances, and all of the mesh currents flow clockwise, we
can enter the matrices directly into MATLAB.
>> Rw = 0.1; Rn=0.1; R1 = 15; R2 = 5; R3 = 8;
>> R = [Rw+Rn+R1 -Rn -R1; -Rn Rw+Rn+R2 -R2; -R1 -R2 R1+R2+R3];
>> V = [120; 120; 0];
>> I = R\V;
>> % Finally, we compute the voltages across the loads.
>> Vr1 = R1*(I(1) - I(3)), Vr2 = R2*(I(2) - I(3)), Vr3 = R3*I(3)
Vr1 =
117.8069
Vr2 =
113.3613
Vr3 =
231.1682
(b) Next, we change Rn to a very high value such as 109 which for
practical calculations is equivalent to an open circuit, and again compute
the voltages resulting in:
Vr1 =
173.9130
Vr2 =
57.9710
Vr3 =
231.8841
The voltage across R1 is certainly high enough to damage most loads
designed to operate at 110 V.
P2.80*
First, we write a node voltage equation to solve for the open-circuit
voltage:
v oc − 10
10
+
v oc
5
=1
Solving, we find v oc = 6.667 V .
Then zeroing the sources, we have this circuit:
Thus, Rt =
1
= 3.333 Ω .
1 10 + 1 5
The Thévenin and Norton equivalents
are:
P2.83
First, we solve the network with a short circuit:
Req = 10 +
1
= 16 Ω
1 10 + 1 15
i10 = 32 Req = 2 A
i15 = i10
10
= 0. 8 A
10 + 15
isc = i15 = 0.8 A
Zeroing the source, we have:
Combining resistances in series and parallel we find Rt = 12 Ω .
Then the Thévenin voltage is vt = isc Rt = 9.6 V . The Thévenin and Norton
equivalents are:
P2.84
The 7-Ω resistor has no effect on the equivalent circuits because the
voltage across the 12-V source is independent of the resistor value.
P2.88
Open-circuit conditions:
ix =
15 − v x
5
vx
10 + 10
− ix + 0.5ix = 0
v x = 10 V and then we have Vt = v oc = v x
Solving, we find
10
= 5 V.
10 + 10
Short-circuit conditions:
ix =
15 − v x
5
vx
10
and then we have isc =
− ix + 0.5ix = 0
vx
10
Thus the equivalents are:
Solving, we find v x = 7.5 V
= 0.75 A. Then, we have Rt = v oc isc = 6.67 Ω .
P2.91*
To maximize the power to RL , we must maximize the voltage across it.
Thus, we need to have Rt = 0 .
Pmax =
P2.94*
The maximum power is
20 2
= 80 W
5
First, we zero the current source and find the current due to the voltage
source.
iv = 30 15 = 2 A
Then, we zero the voltage source and use the current-division principle to
find the current due to the current source.
ic = 3
10
= 2A
5 + 10
Finally, the total current is the sum of the contributions from each source.
i = iv + ic = 4 A
P2.100
We start by assuming i6 = 1 A and work back through the circuit to
determine the value of Is.
This results in Is = -4.5 A.
However, the circuit actually has Is = 10 A, so the actual value ofi6 is
10
× (1 A) = − 2.222 A.
− 4. 5
P2.102
From Equation 2.90, we have
(a)
(b)
Rx =
Rx =
R2
1 kΩ
R3 =
× 3523 = 352.3 Ω
R1
10 kΩ
R2
100 kΩ
R3 =
× 3523 = 35230 kΩ
R1
10 kΩ