Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

Example: 1.6-1
The current through a two-terminal element is given by 
i(t) 
=
10(1 
-
e
-
1000t
) mA for all t 
≥
0 and 
=
0 for t 
≤
0. 
(i) Find the amount of charge that went through the element in 
[0, 5 ms]. (ii) Find an expression for the charge that went through 
the element up to the time instant t. (iii) If the voltage across the 
element is a constant at 10 Volts and the current i(t) flows out 
of positive terminal find and plot the energy delivered by the 
element as a function of time.
Solution
Refer to Fig. 1.6-5. i(t) as per passive sign convention is 
-
-
-
10 1
1000
(
)
e
t
mA.
Fig. 1.6-5 
The two-terminal 
element in
Example: 1.6-1
+
–
v
(
t
) = 10 V
i
(
t
) = –10(1 – e
–1000
t
) mA
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1.34
Circuit Variables and Circuit Elements
(i) Charge that went through the element in a time interval [t
1
, t
2
] is given by 
∆
q
i t dt
t
t
=
∫
( ) .
1
2
Substituting the time-function for i(t) and using limits t
1
=
0 and t
2
=
5 ms, we get,
∆
q
=
−
−
×
= −
+
−
−
−
−
−
∫
10 1
10
10
1
1000
3
0
0 005
2
3
1000
0
0 00
(
)
(
)
.
.
e
dt
t
e
t
t
C
0
55
2
3
5
3
10
0 005 10
10
40
C
C
C.
= −
+
−


= −
−
−
−
−
.
e
m
(ii) The required expression is obtained by
q t
i t dt
i t dt
i t dt
dt
t
t
( )
( )
( )
( )
(
=
=
+
=
+ − ×
× −
−∞
−∞
−∞
−
∫
∫
∫
∫
0
0
0
3
0
10 10
1
e
−−
−
−
−
−
∫
= −
−
−
−
1000
0
5
1000
0
5
1000
0 01 10
0 01 10
t
t
t
t
t
dt
t
e
t
e
)
.
)
.
C
= 
+110 C
C with in ms.
−
−
=
− −


5
10 1 t e
t
t
m
(iii) The instantaneous power delivered to the element is p(t) 
=
v(t) i(t) where v(t) and i(t) are 
as per passive sign convention. Therefore, the power delivered by the element is given by 
– v(t) i(t). The energy delivered by the element is obtained by integrating this quantity as 
below.
Energy delivered
e
= −
= − ×
+ − × −
−
−∞
−
∫
v t i t dt
dt
t
t
( ) ( )
(
10 0
10
10 1
1000
))
.
.
.
.
.
dt
t
e
t
e
t
t
t
mJ
mJ
m
0
0
1000
100
0 1
0 1
0 1
0 1
0 1
∫
∫
−∞
−
−
=
+
−
(
)
=
+
−
(
)
JJ with in ms.
t
Example: 1.6-2
The voltage across a two-terminal element and current through it are given in Fig. 1.6-6. Passive sign 
convention may be assumed. Obtain the instantaneous power delivered to the element and the energy 
delivered to the element as functions of time.
Fig. 1.6-6 
Voltage and current waveform for Example: 1.6-2
v
(
t
)
i
(
t
)
(V)
(A)
(a)
(b)
Time in ms
–3
3
8
6
4
2
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
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PowerandEnergyRelationsforTwo-TerminalElements
 
 
1.35
Solution
Instantaneous power delivered to the element is obtained by p(t) 
=
v(t) i(t). This waveform will 
contain straight-line segments since the current waveform contains straight-line segments and 
voltage waveform is a symmetric rectangular pulse waveform. The power waveform is shown in (a) 
of Fig. 1.6-7.
Fig. 1.6-7 

(a)Waveformofinstantaneouspowerand(b)Waveformofenergy
inExample:1.6-2
p
(
t
)
E
(
t
)
(mJ)
(W)
(a)
(b)
Time in ms
–9
–18
18
9
1 2 3 4 5 6 7 8 9
–18
18
1 2 3 4 5 6 7 8 9
Time in ms
The energy delivered to the element is obtained by integrating the power delivered to the element 
from t 
=
-∞
to t 
=
t. The equation of p(t) in the interval [0, 2 ms] is that of a straight-line of slope 
18 W/ms. Integrating this straight-line equation results in a parabolic curve for energy in that 
interval. The parabolic curve reaches 18 mJ value at 2 ms (since area of the triangle in p(t) curve is 
18 W 
×
2 ms 
×
0.5 
=
18 mJ.) Then p(t) reverses polarity and remains negative and linear in the interval 
[2 ms, 4 ms]. This means that the element delivers power during this interval. The area of triangle in 
the power curve in the interval [2 ms, 4 ms] is again 18 mJ; but with a negative sign. Therefore, the 
total energy delivered to the element at the end of 4 ms period must be 18 mJ 
- 
18 mJ 
=
0 mJ and the 
energy curve between 2 ms and 4 ms must be parabolic again. The variation of energy delivered to
the element is shown in (b) of Fig. 1.6-7.
Note that the net energy delivered to the element at the end of 8 ms is zero. The element received a 
total of 36 mJ of energy during the intervals [0, 2 ms] and [4 ms, 6 ms]. The element delivered a total 
of 36 mJ of energy during the two intervals [2 ms, 4 ms] and [6 ms, 8 ms].
Example: 1.6-3
In charging a storage battery, it is found that energy of 2 watt-hour is expended in 30 minutes in 
sending 200 C through the battery. (i) What is the terminal voltage of the battery assuming that this 
voltage remains constant during the charging process? (ii) What is the magnitude of average charging 
current?
Solution
(i) 200 C of charge went through the battery. Energy delivered to the battery is given by 
v t i t dt
( ) ( ) .
0
30 60
×
∫
The battery voltage is stated to be a constant during the charging process. Let this constant 
voltage be V volts. Then, the energy delivered over 1800 seconds is 
Vi t dt V
i t dt VQ
( )
( )
0
1800
0
1800
∫
∫
=
=
 
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1.36


CircuitVariablesandCircuitElements
where Q is the charge that went through the battery in the same time interval. Therefore, VQ 
=
2 
watt-hour 
=
2 
× 
3600 watt-sec 
=
7200 joules. Since Q is 200 C, V 
=
7200/200 
=
36 Volts.
(ii) The average charging current is the value of a constant current that will result in same charge 
flow over the same time interval. Therefore, the average charging current is 200 C/1800 sec 
=
1/9 Amps.
Example: 1.6-4
Find the current I in the direction marked in Fig. 1.6-8.
Fig. 1.6-8 
CircuitforExample:1.6-4
5 A
5 A
10 V
10 A
20 V
–20 A
–5 V
I
20 A
15 V
15 V
+
+
+
+
+
+
–
–
–
–
–
–
–5 V
Solution
The sum of power delivered by all elements in an isolated circuit must be zero at all instants. Power 
delivered by an element in a DC circuit 
=
–VI where V and I are its voltage and current variables as 
per passive sign convention.
The values of V and I for 10 V source 
=
10 V and –5 A
∴
Power delivered by 10 V source 
=
50 W
The values of V and I for 20 V source 
=
20 V and –20 A
∴
Power delivered by 10 V source 
=
400 W
The values of V and I for 15 V source 
=
15 V and I A
∴
Power delivered by 15 V source 
=
–15I W
The values of V and I for 5 A source 
=
–5 V and 5 A
∴
Power delivered by 5 A source 
=
25 W
The values of V and I for 10 A source 
=
15 V and –10 A
∴
Power delivered by 5 A source 
=
150 W
The values of V and I for –20 A source 
=
–5 V and –20 A
∴
Power delivered by 5 A source 
=
–100 W
Sum of power delivered by all elements 
=
(50 
+ 
400 
-
15I 
+
25 
+ 
150 
-
100) W
=
(525 – 15I) W
This has to be equal to zero. Therefore, the value of I is 35 Amps.