Electric Power and Sign Convention1

7/27/2019 Electric Power and Sign Convention1

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CHAPTER 1:

FUNDAMENTAL OFELECTRIC CIRCUIT


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LEARNING OBJECTIVEApply the passive sign convention and

compute the power dissipated by circuit

elements.

SUBCHAPTER 1.4

ELECTRIC POWER AND SIGN

CONVENTION


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ELECTRIC POWER

Poweris the time rate of expending or absorbing energy,measured in watts (W).

The electric power generated by an active element, or that

dissipated or stored by a passive element, is equal to theproduct of the voltage across the element and the current

flowing through it.

Power = work = work x charge = voltage x current

time charge time

p= dw = dw x dq = v i

d t dq d t

P = V I


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ELECTRIC POWER

In general:

Energy absorbed or supplied by an element from time t0 totime t is:

Energyis the capacity to do work, measured in joules (J).

Power absorbed = Power supplied

t

t

t

t

dtvidtpw

00


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SIGN CONVENTION

Passive sign convention states that:

The power dissipated by a load is a positive quantity

i

i

Power dissipated (-)

Power generated (+)

Power dissipated (+)

Power generated (-)


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STEPS REQUIRED

Choose an arbitrary direction of current flow

Label polarities of all active elements (voltage

and current sources)

Assign polarities to all passive elements

(resistors and load), for passive elements the

current always flow into the positive terminal

Compute the power dissipated by each

element according to the following rule


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STEPS REQUIRED (cont..)

Rules:

If positive current flows into positive terminal of

an element, then the power dissipated is positive (

element absorbs power)

If the current leaves the positive terminal of an

element, then the power dissipated is negative

(element delivers power)


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EXAMPLE (cont..)

Solution:

Known Quantities: Currents through elements D and E; voltage

across elements B, C, E.

Find: Which components are absorbing power, which aresupplying power; verify the conservation of power.

Analysis:

Step 1: Choose an arbitrary direction of current flow.

- Current flow is in clockwise direction

Step 2: Label polarities for voltage or current sources


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EXAMPLE 1 (cont..)

Therefore,

- By KCL, the current through element B is 5 A, to the

right.

- By KVL: -va

3 +10 + 5 = 0

Therefore, the voltage across element A is:

va = 12 V (positive at the top)


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EXAMPLE 1 (cont..)

Step 3: Compute the power dissipated by each elementA supplies (12 V)(5 A) = 60 W

B supplies (3 V)(5 A) = 15 W

C absorbs (5 V)(5 A) = 25 W

D absorbs (10 V)(3 A) = 30 W

E absorbs (10 V)(2 A) = 20 W

Total power supplied = 60 W + 15 W = 75 W

Total power absorbed = 25 W+30W+20W= 75 W

Total power supplied = Total power absorbed

so, conservation of power is satisfied.


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Thank You

Electric Power and Sign Convention1

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