Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

i
x
10 
Ω
2 A
0.2 H
0.5 H
3 A
10 
Ω
5 
Ω
Fig. 2.9-13 
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Problems 
2.39
14. Find the currents in all the resistors in the circuit shown in Fig. 2.9-14 by applying Kirchhoff’s 
laws with I
1
=
2 A, I
2
=
5 A and I
3
=
2 A. [Hint: Write KCL equations at three nodes A, B and C in 
terms of voltage variables V
AD
, V
BD
and V
CD
and resistance values.] 
5 
Ω
A
B
C
D
5 
Ω
5 
Ω
2 
Ω
2 
Ω
1 
Ω
I
2
R
2
I
3
I
1
Fig. 2.9-14 
15. Solve the circuit shown in Fig. 2.9-15 and find the power consumed by the resistors, power 
delivered by the independent source and power delivered by the dependent source.
10 
Ω
5 
Ω
10 V
– 0.2 
v
x
v
x
+
+
+
–
–
–
Fig. 2.9-15 
16. Solve the circuit shown in Fig. 2.9-16 completely and find out V
AB
. Also find the total power 
dissipated in the circuit and the power delivered by independent and dependent sources.
+
+
+
+
–
–
–
–
10 
Ω
5 
Ω
v
x
– 0.5 
v
x
10 V
B
A
Fig. 2.9-16 
17. Find the circuit current and power delivered by all the six elements in the circuit shown in
Fig. 2.9-17.
0.5 
v
x
v
x
2 
i
x
i
x
6 
Ω
3 
Ω
2 
Ω
13 V
+
+
+
+
–
–
–
–
Fig. 2.9-17 
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2.40
Basic Circuit Laws
18. Find the voltage across the parallel combination in the circuit shown in Fig. 2.9-18. Also find the 
power absorbed by all the elements in the circuit.
i
x
i
x
v
x
v
x
3 
Ω
4 
Ω
2 
Ω
2 
Ω
0.5
0.75
3 A
+
–
Fig. 2.9-18 
19. Find the energy delivered to the 9 
W
resistor and dissipated in the 2 
W
resistors in the circuit 
shown in Fig. 2.9-19 during [0, 2 s] if v
s1
=
10 sin100
p
t V and v
s2
=
10 V.
2 
Ω
2 
Ω
v
S1
+
–
+
–
Fig. 2.9-19 
20. Find the charge delivered to the 6 V voltage source from t 
=
0 to t 
=
2 s in the circuit in Fig. 2.9-20. 
i
s1
=
2
+
e
-
 
t
A for t 
≥
0 and 0 A for t < 0. i
s2
=
te
-
2t
A for t 
≥
0 and 0 A for t < 0.
i
S1
i
S2
10 
Ω
6 V
+
–
Fig. 2.9-20 
21. Find the coefficients k
1
and k
2 
for the dependent sources in the circuit in Fig. 2.9-21.
k
1
i
X
i
X
k
2
v
y
v
y
0.2 
Ω
0.2 
Ω
1 
Ω
1 
Ω
0.5 
Ω
0.5 
Ω
1 A
4 A
4 A
15 A
2 A
+
–
+
–
Fig. 2.9-21 
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Problems 
2.41
22. Find 
a
and 
b
in the circuit in Fig. 2.9-22.
v
x
α
i
x
β
v
x
i
x
4 
Ω
1 
Ω
1 
Ω
2 V
3 V
1 V
3 V
3 
Ω
2 
Ω
+
+
+
+
+
+
–
–
–
–
–
–
V
1
Fig. 2.9-22 
23. Design an Opamp circuit to produce v
o
(t) 
=
3 
+
7 sin 200
p
t V using 
±
12 V power supply and a 
signal source v
S
(t) 
=
0.5 sin 200
p
t V.
24. Find i
L
, i
s
, i
o
and v
o 
in the voltage to current converter in Fig. 2.9-23 designed to produce a 
constant current in a 10 
W
load.
1080 
Ω
1080 
Ω
120 
Ω
120 
Ω
10 
Ω
Load
12 V
i
o
v
o
i
L
R
L
i
S
R
4
R
2
R
1
R
3
+
+
–
+
–
–
Fig. 2.9-23
25. (i) Show that the current in the load resistance R
L
in the circuit in Fig. 2.9.24 is independent of R
L
if (R
3
+
R
4
) >> R
5
//R
L
. (ii) Find i
L
, i
s
, i
o
and v
o 
with the component values shown.
100 k
Ω
100 k
Ω
0.5 k
Ω
1 k
Ω
1 k
Ω
10 
Ω
Load
i
o
v
o
i
L
R
L
R
4
R
2
R
1
+
+
–
–
12 V
i
S
R
3
+
–
Fig. 2.9.24
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2.42
Basic Circuit Laws
26. Show that v
o
(t) 
=
(1 
+ 
R
2
/R
1
)(v
1
-
v
2
) in the circuit in Fig. 2.9-25.
v
o
v
2
v
1
R
1
R
1
R
2
R
2
+
–
+
–
+
–
Fig. 2.9-25
27. (i) Show that the overall feedback is of degenerative nature in the circuit in Fig. 2.9-26. (ii) Derive 
expressions for gains at the outputs of both Opamps. (iii) Evaluate the gain v
o
/v
s
 with R
4
=
99R, 
R
3
 
=
100R and R
2
 
= 
9R
1
. (iv) Design the circuit using 
m
A741 IC with the above constraints on 
resistances. 
v
o
v
S
+
–
–
+
R
1
R
2
R
3
R
4
R
R
Fig. 2.9-26
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S i n g l e E l e m e n t 
C i r c u i t s
CHAPTER OBJECTIVES
• Voltage–current relation of a resistor.
• Voltage, current and power division principle in series and parallel resistor combinations.
• Voltage–current relation of an Inductor and its various implications.
• Initial current in an inductor and its significance.
• Series and parallel combination of inductors.
• Voltage, current and power sharing in series and parallel connection of inductors.
• Voltage–current relation of a capacitor and its various implications.
• Initial voltage across a capacitor and its significance.
• Series and parallel combination of capacitors.
• Voltage, current and power sharing in series and parallel connection of capacitors.
• Simple single element circuits with independent voltage source and current source excitation.
• Unit impulse function and unit step function.
IntroductIon
In this chapter, we study simple circuits containing one type of element – resistor, inductor or
capacitor – driven by one source, either an independent voltage source or independent current source. 
These circuits may contain more than one element, but all of them will be of same type except the 
source. The aim of this study will be to understand the behaviour of each element type thoroughly. 
We will also deal with ‘series and parallel equivalents’ that can be used to replace series or parallel 
connection of multiple elements of same type by one equivalent element of that type. Moreover, we 
will meet with two very interesting source functions – unit impulse function 
d
 (t) and unit step function 
u(t). These functions are extremely important in Circuit Analysis. 
Chapter 
3
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3.2
Single Element Circuits
3.1 
the resIstor
The physical basis for the two-terminal resistor was dealt with in detail in Chapter 1. The graphic 
symbol of a linear resistor and its element relationship is shown in the following:
i
(
t
)
v
(
t
)
+
–
R
v t
Ri t
i t
G v t
t
p t
v t i t
R i t
v t
( )
( )
( )
( )
( )
( ) ( )
( )
( )
=
=
=
=
[ ]
=
[ ]
or
for all
2
22
2
2
R
i t
G
G v t
=
[ ]
=
[ ]
( )
( )
where p(t) is the power delivered to the resistor in watts, R is the resistance 
of the resistor and G is the conductance of the resistor. R and G are reciprocals of each other. 
The current response in a resistor at a particular instant depends only on the voltage applied 
across it at that instant. Therefore, a resistor is a memoryless element. The waveshape of voltage 
and current have to be the same in such an element. A resistor cannot change the waveshape of a
signal.
A resistor can only dissipate energy. Therefore, the power delivered to a positive resistor is always 
positive or zero.