Full file at http://testbank360.eu/solution-manual-the-analysis-and-design-of-linear-circuits-6th-edition-thomas
Problem 2-1
The current through a 33-kΩ resistor is 1.2 mA. Find the voltage across the resistor.
Solution:
v = iR
clear all
R = 33e3;
ii = 1.2e-3;
v = ii*R
v =
39.6000e+000
Answer:
v = 39.6 V
Problem 2-2
A 6.2-kΩ resistor dissipates 12 mW. Find the current through the resistor.
Solution:
p = i2R
clear all
format short eng
R = 6.2e3;
p = 12e-3;
ii = sqrt(p/R)
ii =
1.3912e-003
Answer:
i = ±1.3912 mA
Problem 2-3
The conductance of a particular resistor is 1 mS. Find the current through the resistor when connected across a 9 V source.
Solution:
v = iR
clear all
G = 1e-3;
R = 1/G;
v = 9;
ii = v/R
ii =
9.0000e-003
Answer:
i = 9 mA
Problem 2-4
In Figure P2-4 the resistor dissipates 25 mW. Find Rx.
Solution:
clear all
p = 25e-3;
v = 15;
R = v^2/p
R =
9.0000e+003
Answer:
R = 9 kΩ
Problem 2-5
In Figure P2-5 find Rx and the power delivered to the resistor.
Solution:
clear all
v = 100;
ii = 10e-3;
R = v/ii
p = v*ii
R =
10.0000e+003
p =
1.0000e+000
Answer:
Rx = 10 kΩ, p = 1 W
Problem 2-6
The i-v characteristic of a nonlinear resistor are v = 75i + 0.2i3.
(a) Calculate v and p for i = ±0.5, ±1, ±2, ±5, and ±10 A.
(b) Find the maximum error in v when the device is treated as a 75-Ω linear resistance on the range |i| < 0.5 A.
Solution:
clear all
format short eng
ii = [-10, -5, -2, -1, -0.5, 0.5, 1, 2, 5, 10];
v = 75*ii + 0.2*ii.^3;
p = v.*ii;
Results = [ii' v' p']
syms i1
v1 = 75*i1+0.2*i1^3;
v2 = 75*i1;
ii1 = -0.5:0.01:0.5;
vv1 = subs(v1,i1,ii1);
vv2 = subs(v2,i1,ii1);
plot(vv1,ii1,'b','LineWidth',3)
hold on
plot(vv2,ii1,'g','LineWidth',1)
grid on
xlabel('Voltage (V)')
ylabel('Current (A)')
legend('Nonlinear','Linear')
MaxError = max(vv1)-max(vv2)
MaxError2 = subs(v1-v2,i1,0.5)
Results =
-10.0000e+000 -950.0000e+000 9.5000e+003
-5.0000e+000 -400.0000e+000 2.0000e+003
-2.0000e+000 -151.6000e+000 303.2000e+000
-1.0000e+000 -75.2000e+000 75.2000e+000
-500.0000e-003 -37.5250e+000 18.7625e+000
500.0000e-003 37.5250e+000 18.7625e+000
1.0000e+000 75.2000e+000 75.2000e+000
2.0000e+000 151.6000e+000 303.2000e+000
5.0000e+000 400.0000e+000 2.0000e+003
10.0000e+000 950.0000e+000 9.5000e+003
MaxError =
25.0000e-003
MaxError2 =
25.0000e-003
Answer:
(a)
i (A) / v (V) / p (W)-10 / -950 / 9500
-5 / -400 / 2000
-2 / -151.6 / 303.2
-1 / -75.2 / 75.2
-0.5 / -37.525 / 18.7625
0.5 / 37.525 / 18.7625
1 / 75.2 / 75.2
2 / 151.6 / 303.2
5 / 400 / 2000
10 / 950 / 9500
(b) ERRORMAX = 25 mV
Problem 2-7
A 10-kΩ resistor has a power rating of ⅛W. Find the maximum voltage that can be applied to the resistor.
Solution:
clear all
R = 10e3;
p = 1/8;
v_max = sqrt(p*R)
v_max =
35.3553e+000
Answer:
vmax = 35.36 V
Problem 2-8
A certain type of film resistor is available with resistance values between 10 Ω and 100 MΩ. The maximum ratings for all resistors of this type are 500 V and 1/4 W. Show that the voltage rating is the controlling limit for R > 1 MΩ, and that the power rating is the controlling limit when R < 1 MΩ.
Solution:
clear all
V = 500;
p = 1/4;
R = V^2/p
R =
1.0000e+006
At R = 1 MΩ, both p and v can take their maximum values and there are no issues. For R > 1MΩ, with a maximum voltage, the power must be less than 0.25 W, so the voltage rating on a particular resistor will control the maximum allowable value for the power. For R < 1 MΩ, with a maximum voltage, the power will be greater than 0.25 W, so the power rating on a particular resistor will control the maximum allowable value for the voltage.
Answer:
Presented above.
Problem 2-9
Figure P2-9 shows the circuit symbol for a class of two-terminal devices called diodes. The i-v relationship for a specific pn junction diode is
(a) Use this equation to find i and p for v = 0, ±0.1, ±0.2, ±0.4, and ±0.8 V. Use these data to plot the i-v characteristic of the element.
(b) Is the diode linear or nonlinear, bilateral or nonbilateral, and active or passive?
(c) Use the diode model to predict i and p for v = 5 V. Do you think the model applies to voltages in this range? Explain.
(d) Repeat (c) for v = –5 V.
Solution:
clear all
v = [-0.8, -0.4, -0.2, -0.1, 0 0.1, 0.2, 0.4, 0.8];
ii = 2e-16*(exp(40*v)-1);
p = v.*ii;
Results = [v' ii' p']
plot(v,ii,'b','LineWidth',3)
xlabel('Voltage (V)')
ylabel('Current (A)')
grid on
v = 5
i5 = 2e-16*(exp(40*v)-1)
v = -5
iNeg5 = 2e-16*(exp(40*v)-1)
Results =
-800.0000e-003 -200.0000e-018 160.0000e-018
-400.0000e-003 -200.0000e-018 80.0000e-018
-200.0000e-003 -199.9329e-018 39.9866e-018
-100.0000e-003 -196.3369e-018 19.6337e-018
0.0000e-003 0.0000e-003 0.0000e-003
100.0000e-003 10.7196e-015 1.0720e-015
200.0000e-003 595.9916e-015 119.1983e-015
400.0000e-003 1.7772e-009 710.8888e-012
800.0000e-003 15.7926e-003 12.6341e-003
v =
5.0000e+000
i5 =
144.5195e+069
v =
-5.0000e+000
iNeg5 =
-200.0000e-018
Answer:
(a)
v (V) / i (A) / p (W)-0.8 / -2.00E-16 / 1.60E-16
-0.4 / -2.00E-16 / 8.00E-17
-0.2 / -2.00E-16 / 4.00E-17
-0.1 / -1.96E-16 / 1.96E-17
0 / 0 / 0
0.1 / 1.07E-14 / 1.07E-15
0.2 / 5.96E-13 / 1.19E-13
0.4 / 1.78E-09 / 7.11E-10
0.8 / 1.58E-02 / 1.26E-02
(b) The plot in Part (a) shows that the device is nonlinear and nonbilateral. The power for the device is always positive, so it is passive.
(c) For v = 5 V, i = 1.45 ´ 1071 A and p = 7.23 ´ 1071 W. The model is not valid because the current and power are too large.
(d) For v = -5 V, i = −2.00 ´ 10−16 A and p = 1.00 ´ 10−15 W. The model is valid because the current and power are both essentially zero.
Problem 2-10
In Figure P2-10 i2 = –2 A and i3 = 5 A. Find i1 and i4.
Solution:
Apply KCL at Nodes B and C.
clear all
i2 = -2;
i3 = 5;
i1 = -i2
i4 = i2+i3
i1 =
2.0000e+000
i4 =
3.0000e+000
Answer:
i1 = 2 A and i4 = 3 A.
Problem 2-11
For the circuit in Figure P2-11:
(a) Identify the nodes and at least two loops.
(b) Identify any elements connected in series or in parallel.
(c) Write KCL and KVL connection equations for the circuit.
Solution:
There are three nodes and three loops.
Answer:
(a) nodes: A, B, C; loops: 1-2; 2-3-4; 1-3-4
(b) series: 3 and 4; parallel: 1 and 2
(c) KCL: node A: ;
node B: ;
node C:
KVL: loop 1-2: ;
loop 2-3-4: ;
loop 1-3-4:
Problem 2-12
In Figure P2-11, i2 = –10 mA and i4 = 20 mA. Find i1 and i3.
Solution:
Use the KCL equations developed in the solution to Problem 2-11.
clear all
i2 = -10e-3;
i4 = 20e-3;
i3 = i4
i1 = -i2-i3
i3 =
20.0000e-003
i1 =
-10.0000e-003
Answer:
i1 = -10 mA and i3 = 20 mA.
Problem 2-13
For the circuit in Figure P213:
(a) Identify the nodes and at least three loops in the circuit.
(b) Identify any elements connected in series or in parallel.
(c) Write KCL and KVL connection equations for the circuit.
Solution:
There are four nodes and at least five loops. There are only three independent KVL equations.
Answer:
(a) nodes: A, B, C, D;
loops: 1-3-2; 2-4-5; 3-6-4; 1-6-5; 2-3-6-5; 1-6-4-2; 1-3-4-5
(b) series: none; parallel: none
(c) KCL: node A: ;
node B: ;
node C: ;
node D:
KVL: loop 1-3-2: ;
loop 2-4-5: ;
loop 3-6-4:
Problem 2-14
In Figure P2-13 v2 = 10 V, v3 = –10 V, and v4 = 3 V. Find v1, v5, and v6.
Solution:
Use the KVL equations developed in the solution to Problem 2-13.
clear all
v2 = 10;
v3 = -10;
v4 = 3;
v1 = v2 - v3
v5 = v2 - v4
v6 = v4 - v3
v1 =
20.0000e+000
v5 =
7.0000e+000
v6 =
13.0000e+000
Answer:
v1 = 20 V, v5 = 7 V, and v6 = 13 V.
Problem 2-15
The circuit in Figure P2-15 is organized around the three signal lines A, B, and C.
(a) Identify the nodes and at least three loops in the circuit.
(b) Write KCL connection equations for the circuit.
(c) If i1 = –20 mA, i2 = –12 mA, and i3 = 50 mA, find i4, i5, and i6
(d) Show that the circuit in Figure P2-15 is identical to that in Figure P2-13.
Solution:
(a) There are four nodes and at least five loops.
(b) KCL: node A: ;
node B: ;
node C: ;
node D:
clear all
i1 = -20e-3;
i2 = -12e-3;
i3 = 50e-3;
i4 = -i2-i3
i5 = -i1-i2
i6 = i3-i1
i4 =
-38.0000e-003
i5 =
32.0000e-003
i6 =
70.0000e-003
Answer:
(a) nodes: A, B, C, D;
loops: 1-3-2; 2-4-5; 3-6-4; 1-6-5; 2-3-6-5; 1-6-4-2; 1-3-4-5
(b) KCL: node A: ;
node B: ;
node C: ;
node D:
(c) i4 = −38 mA; i5 = 32 mA; i6 = 70 mA
(d) The circuits have the same nodes, connections, and current directions, so they must be equivalent.
Problem 2-16
In Figure P2-16, v2 = 10 V, v3 = 10 V, and v4 = 10 V. Find v1 and v5.
Solution:
Apply KVL to the circuit.
clear all
v2 = 10;
v3 = 10;
v4 = 10;
v1 = v2+v3
v5 = v3-v4
v1 =
20.0000e+000
v5 =
0.0000e-003
Answer:
v1 = 20 V and v5 = 0 V.
Problem 2-17
In Figure P2-17 i2 = 10 mA, i3 = –15 mA, and i4 = 5 mA. Find i1 and i5.
Solution:
Apply KCL to the circuit.
clear all
i2 = 10e-3;
i3 = -15e-3;
i4 = 5e-3;
i1 = i2-i3+i4
i5 = i2-i1
i1 =
30.0000e-003
i5 =
-20.0000e-003
Answer:
i1 = 30 mA and i5 = −20 mA
Problem 2-18
(a) Use the passive sign convention to assign voltage variables consistent with the currents in Figure P2-17. Write three KVL connection equations using these voltage variables.
(b) If v3 = 0 V, what can be said about the voltages across all the other elements?
Solution:
(a) Voltage signs:
Elements 1 and 3: plus on bottom and minus on top
Elements 2 and 4: plus on top and minus on bottom
Element 5: plus on left and minus on right
Write the KVL equations for the loops formed by 1-2, 3-4, and 2-4-5
loop 1-2:
loop 3-4:
loop 2-4-5:
(b) If v3 = 0 V, then v4 = 0 V. In addition, v2 = -v5 and v1 = v5.
Answer:
Presented above.
Problem 2-19
The KCL equations for a three-node circuit are:
Node A – i1 + i2 – i4 = 0
Node B – i2 – i3 + i5 = 0
Node C i1 + i3 + i4 – i5 = 0
Draw the circuit diagram and indicate the reference directions for the element currents.
Answer:
Problem 2-20
Find vx and ix in Figure P2-20.
Solution:
Use KCL to find the current and Ohm's Law to find the voltage.
clear all
format short eng
is = 2e-3;
ix = -is
vx = 47e3*ix
ix =
-2.0000e-003
vx =
-94.0000e+000
Answer:
vx = -94 V and ix = -2 mA.
Problem 2-21
Find vx and ix in Figure P2-21.
Solution:
Find the voltage across the 10-Ω resistor using Ohm's Law. The 10-Ω and 5-Ω resistors are in parallel, so they have the same voltage. Find the current through the 5-Ω resistor. The current through the 4-Ω resistor is the sum of the currents through the other two resistors. Find the voltage across the 4-Ω resistor. Then vx is the sum of the voltages across the 4-Ω and 10-Ω resistors.
clear all
i10 = 1/2;
v10 = 10*i10;
v5 = v10;
i5 = v5/5;
ix = i5
i4 = i10+i5;
v4 = 4*i4;
vx = v4+v10
ix =
1.0000e+000
vx =
11.0000e+000
Answer:
vx = 11 V and ix = 1 A
Problem 2-22
In Figure P 2-22:
(a) Assign a voltage and current variable to every element.
(b) Use KVL to find the voltage across each resistor.
(c) Use Ohm's law to find the current through each resistor.
(d) Use KCL to find the current through each voltage source.
Solution:
(a) For each of the three resistors, the voltage positive sign is on the left and the negative sign is on the right. The current flows from left to right through each element.
Element 1: 50-Ω resistor.
Element 2: left 100-Ω resistor.
Element 3: right 100-Ω resistor.
The left voltage source is vS1, with iS1 flowing down.
The center voltage source is vS2, with iS2 flowing down.
The right voltage source is vS3, with iS3 flowing down.
(b) KVL equations:
clear all
format short eng
vs1 = 5;
vs2 = 10;
vs3 = 5;
v1 = vs1-vs3
v2 = vs1-vs2
v3 = vs2-vs3
v1 =
0.0000e-003
v2 =
-5.0000e+000
v3 =
5.0000e+000
(c) v = iR
i1 = v1/50
i2 = v2/100
i3 = v3/100
i1 =
0.0000e-003
i2 =
-50.0000e-003
i3 =
50.0000e-003
(d) KCL equations
is1 = -i1-i2
is2 = i2-i3
is3 = i1+i3
is1 =
50.0000e-003
is2 =
-100.0000e-003
is3 =
50.0000e-003
Answer:
(a) Presented above.
(b) v1 = 0 V, v2 = -5 V, and v3 = 5 V
(c) il = 0 mA, i2 = -50 mA, and i3 = 50 mA
(d) iS1 = 50 mA, iS2 = -100 mA, and iS3 = 50 mA
Problem 2-23
Find the power dissipated in the 1.5 kΩ resistor in Figure P2-23.
Solution:
Label the elements.
Element 1: 1-kΩ resistor with current flowing down
Element 2: 500-Ω resistor with current flowing to the right
Element 3: 1.5-kΩ resistor with current flowing down
Write KCL, KVL, and Ohm's Law equations:
i1 + i2 - 5 mA = 0
-i2 + i3 = 0