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POWER
(Received/ Supplied)
PASSIVE SIGN CONVENTION
POWER RECEIVED IS POSITIVE WHILE POWER
SUPPLIED IS CONSIDERED NEGATIVE
+ Vab −
a
b
I ab
P = Vab I ab
IF VOLTAGE AND CURRENT
ARE BOTH POSITIVE THE
CHARGES MOVE FROM
HIGH TO LOW VOLTAGE
AND THE COMPONENT
RECEIVES ENERGY --IT
IT IS
A PASSIVE ELEMENT
A CONSEQUENCE OF THIS CONVENTION IS THAT
THE REFERENCE DIRECTIONS FOR CURRENT AND
VOLTAGE ARE NOT INDEPENDENT -- IF WE
ASSUME PASSIVE ELEMENTS
GIVEN THE REFERENCE POLARITY
+ Vab −
a
b
REFERENCE DIRECTION FOR CURRENT
THIS IS THE REFERENCE FOR POLARITY
+
−
a
b
I ab
IF THE REFERENCE DIRECTION FOR CURRENT
IS GIVEN
EXAMPLE
+ Vab −
2A
a
b
I ab
Vab = −10V
THE ELEMENT RECEIVES 20W OF POWER.
WHAT IS THE CURRENT?
SELECT REFERENCE DIRECTION BASED ON
PASSIVE
SS
S
SIGN
G CO
CONVENTION
O
20[W ] = Vab I ab = ( −10V ) I ab
I ab = −2[ A]
UNDERSTANDING PASSIVE SIGN CONVENTION
We must examine the voltage across the component
and the current through it
I
A
+
V
S1
B
Current A - A'
positive
iti
positive
iti
positive negative
negative positive
negative negative
Voltage(V)
A’
−
S1
supplies
li
receives
receives
supplies
PS1 = V AB I AB
S2
PS 2 = V A'B ' I A'B '
B’
S2
ON S1
ON S2
receives
i
V AB > 0, I AB < 0 V A B > 0, I A B > 0
supplies
ON S2
supplies
V A'B ' < 0, I A'B ' > 0
receives
'
'
'
'
DETERMINE WHETHER THE ELEMENTS ARE SUPPLYING OR RECEIVING POWER
AND HOW MUCH
a
a
Vab = 2V
I ab = 4 A
2A
I ab = 2 A
Vab = −2V
P = −8W
SUPPLIES POWER
b
P = 4W
RECEIVES POWER
1
1
2
2
V12 = 12V , I12 = −4 A
V12 = 4V , I12 = 2 A
b
I = −8[ A]
V AB = −4[V ]
− 20[W ] = V AB × (5 A)
+
+
−
−
SELECT VOLTAGE REFERENCE POLARITY
BASED ON CURRENT REFERENCE DIRECTION
40[W ] = (−5V ) × I
WHICH TERMINAL HAS HIGHER VOLTAGE AND WHICH IS THE CURRENT FLOW DIRECTION
V1 = −20[V ]
40[W ] = V1 × ( −2 A)
− 2A
I = −5[ A]
SELECT HERE THE CURRENT REFERENCE DIRECTION
− 50[W ] = (10[V ]) × I
BASED ON VOLTAGE REFERENCE POLARITY
COMPUTE POWER ABDORBED OR SUPPLIED BY EACH ELEMENT
P1 = (6V )(2 A)
2 A + 6V −
+
24V
−
+
1
+
-
3
2
2A
18V
−
P1 = 12W
P2 = 36W
P3 = -48W
P2 = (18V )(2 A)
P3 = (24V )(−2 A) = (−24V )(2 A)
IMPORTANT: NOTICE THE POWER BALANCE IN THE CIRCUIT
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