BEF 12403 - Week 2 - Power and Energy.ppt

Power
v
i = dq/dt
Let us consider now the rate at which energy is being delivered to, or by, a
circuit element by a current i(t).
If the voltage across the element is v and a small charge
q is moved through the element from the positive to the
negative terminal, the energy absorbed by the element
(w, say) is given by
q
v
w 


If the time involved is t, then the rate at which work is being done is given
by the ratio w/t. Thus, dividing both sides of the above equation by t,
we obtain
t
q
v
t
q
v
t
w









v
i = dq/dt
In the limit that t→0, we obtain














 t
q
v
t
w
t
t 0
0
lim
lim
or







dt
dq
v
dt
dw
or
Since by definition the rate at which energy is expended is power,
denoted by p, we have
vi
dt
dq
v
dt
dw
p 









The quantities v and i are generally functions of time, which
we may also denote by v(t) and i(t). Therefore, p in the
above expression is a time-varying quantity. It is sometimes
called the instantaneous power because its value is the
power at the instant of time at which v and i are measured.
Hence
p(t) = v(t)i(t)
p = vi
To explicitly show the time dependence we sometimes write

Measuring power
One way of measuring the power consumed by a
circuit element is diagrammed in Figure x. We
connect a current-sensing instrument called an
ammeter so that the current i flows through both
the element and the meter. We also connect a
voltmeter to read the voltage v across the element.
The product of the two meter readings then equals
the power – providing the voltmeter is ideal so that
all of the current i measured by the ammeter
passes through the element.
For convenience we simplify the circuit drawing as
shown in Figure 1.2b.
Ammeter
Voltmeter
v
i

Passive Sign Convention
We must consider reference directions when we
write a power formula. The sign in the power
formula depends on the combination of the voltage
and current reference directions.
For the voltage and current reference directions
shown in Figure x, where the current arrow is
shown directed into the + polarity marking of the
voltage, the power absorbed by the element is
computed using the expression
vi
p 
Using the given current and voltage reference directions with the
expression, we can deduce that the element consumes power if
p is +ve and generates power if p is negative.
v
i

Passive Sign Convention (continued)
On the other hand, if we want to compute the
power generated by the element, then we use the
expression
vi
p 

Using the given reference directions and the above power
expression, we can deduce that the element generates power if
p is +ve and consumes power if p is negative.
v
i

Active Sign Convention
The active sign convention is used when the
current reference direction is directed out of the +
polarity marking of the voltage.
v
i
Power absorbed (or consumed) element is
computed from the expression
p = - vi
Here power is consumed if p is positive and is generated if p
is negative.

Active Sign Convention (continued)
v
i
To compute the power generated by the element
we use the expression
p = + vi
Here power is generated if p is positive and is consumed
if p is negative.

Example
Compute the power delivered by each of the following elements.
- 4 V
10 V
2 A
6 V
5 A 1 A
(a) (b) (c)

Solution
(a) Here the 2 A current enters the element via its + terminal.
2 A
6 V
(a)
Hence, power delivered by the element (to the rest of the circuit) is
p = - vi = - (6)(2) = - 12 W
(Since p is negative, power is in fact consumed by the element.)

Solution
(b) Here the 2 A current enters the element via its + terminal.
For this case the 5 A current enters the element via its -
terminal; hence, power delivered by the element (to the rest of
the circuit) is
10 V
5 A
(b)
p = vi = (10)(5) = 50 W
(Since p is positive, power is generated by the element.)

(c) Here the 1 A current enters the element via its – terminal;
hence, power delivered by the element (to the rest of the
circuit) is
p = vi = (-4)(1) = - 4 W
(Since p is negative, the element is generating power.)
- 4 V
1 A
(c)
Solution (continued)

Example
Two electrical devices are connected at an interface, as shown in Figure
x. Using the reference marks shown in the figure, find the power
transferred and state whether the power is transferred from A to B or
from B to A when
v = + 12 V, i = - 2 A;
v = - 33 V, i = -1 A;
v = + 15 V, i = + 4 A;
v = -30 V, i = - 4 A
A B
v
i

Solution
Let us compute the power generated by each device.
For device A, the power generated is:
p = vi = (12)(-2) = - 24 W (absorbed)
p = vi = (-33)(-1) = 33 W (generated)
p = vi = (15)(4) = 60 W (generated)
p = vi = (-30)(-4) = 120 W (generated)
A B
v
i

A B
v
i
For device B, the power generated is:
p = -vi = -(12)(-2) = 24 W (generated)
p = -vi = -(-33)(-1) = -33 W (absorbed)
p = -vi = -(15)(4) = 60 W (absorbed)
p = -vi = -(-30)(-4) = -120 W (absorbed)

Example
Two elements are connected in series as shown in Figure x. Element 1
supplies 24 W of power. Is element 2 supplying or absorbing power, and
how much?
6 V
8 V
1
2

Solution
First we need to find the reference current
direction. To do this we note that since element 1 is
a source, the reference current flows out from its
(+) terminal. The magnitude of the reference
current is
A
4
6
24



v
p
i
Based on the established direction of the reference
currrent, element 2 has the load set reference
directions. Hence, power consumed by element 2 is
6 V
8 V
1
2
W
32
)
4
)(
8
( 

 vi
p
Since p is positive, element 2 is absorbing power.

Energy delivered by a current to an element from time - ∞ to t.
The incremental energy Δw delivered to an element between time to and t
by moving a charge Δq over a potential diffence v (see Figure x) is given by
the expression
vidt
dw 
The total energy delivered from time - ∞ to t is obtained
by integrating the energy over the time interval. Thus,

 




t
t
vid
dw 
v
i = dq/dt
Energy

Energy delivered to an element from time - ∞ to t. (continued)
v
i = dq/dt


 








t
t
vid
vid
vid
w
t
w
0
0
)
(
)
( 


We have, upon integrating both sides between - and t,

v
i = dq/dt
Using the fact the energy delivered in the beginning of time
was zero; that is,
0
)
( 

w
Breaking the integration into two time segments, we
can write


 





t
t
vid
vid
vid
0
0




v
i = dq/dt
Writing



0
)
0
( 
vid
w
thus, we obtain



t
vid
w
t
w
0
)
0
(
)
( 
Total energy over a time interval is found by integrating power



t
pd
w
t
w
0
)
0
(
)
( 

v
i = dq/dt
For the special case where p has the constant (or average)
value P, that is p(t) = P, we can write w(t) = W (say) to
obtain
PT
W 
whose units are watt-seconds or joules.
1 kWh = 3.6 x 106 J
Utility bills are commonly expressed in terms of the
kilowatthour (kWh),
which equals the total energy delivered in one hour when P = 1000 W.

Example
The current and voltage at the terminals of the device shown in Figure Xa
are as sketched in Figures Xba and Xc, respectively.
a) Sketch the power versus time plot;
b) Sketch the energy versus time plot
v
i
t (s)
i (A)
150
2 μs
t (s)
v (volts)
300
2 μs
0
0

Solution
v
i
t (s)
i (A)
150
2 μs
t (s)
v (volts)
300
2 μs
0
0
The voltage and current be expressed as
piecewise functions v(t) and i(t), respectively, as
follows:
otherwise
s
2
0
0
10
5
.
1
)
(
8






t
t
v
otherwise
s
2
0
0
10
75
.
0
)
(
8





t
t
i

v
i
t (s)
i (A)
150
2 μs
t (s)
v (volts)
300
2 μs
0
0
a. The instantaneous power absorbed by the
element is
)
(
)
(
)
( t
i
t
v
t
p 
Hence, the power versus time plot is:
0 5 10
7
1 10
6
1.5 10
6
2 10
6
3750
7500
1.13 10
4
1.5 10
4
p t
( )
t

v
i
t (s)
i (A)
150
2 μs
t (s)
v (volts)
300
2 μs
0
0
b. The total energy absorbed up to time t = 2s is


t
d
p
t
e
0
)
(
)
( 

Hence, the energy versus time plot is:
0 5 10
7
1 10
6
1.5 10
6
2 10
6
0.0037
0.0075
0.0113
0.015
0.015
0
e t
( )
2 10
6


0 t

BEF 12403 - Week 2 - Power and Energy.ppt

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BEF 12403 - Week 2 - Power and Energy.ppt