POWER

Rate of doing work is called power.

•   The unit of electric power is watt.

•   One watt of power is equals the work done in one second by one volt of potential difference in moving one coulomb of charge.

•   We know that one coulomb per second is an ampere. Therefore power in watts equals the product of volt times amperes.

•   Power in watts = volts x amperes

•                                           P = V X I

•                 Dimensionally, the right side of this equation is the product of joules per              coulomb and coulombs per second, which produces the expected dimension                  of joule per second or watt.

•   The sketch shows that if one terminal of the element is v volts positive with respect to the other terminal, and of current i is entering the element through the terminal then the power is absorbed by the element.

•   It is also correct to say that a power   p =vi is being delivered to the element.

PASSIVE SIGN CONVENTIONS

•   If the current arrow is directed into the + marked terminal of an element, then p=vi yields the absorbed power.

•   A negative value indicates the power is actually being generated by the element, it might have been better to define a current flowing out of the + terminal.

Example

Q.  A simple circuit is formed using a12V lead-acid battery and an automobile headlight. If the battery delivers a total energy of 460.8watt-hours over an 8 hours discharge period.

(a)  How much power is delivered to the headlight?

(b)  What is the current flowing through the bulb (assume the battery voltage remains constant while discharging)

      Solution:

      The battery delivers energy of 460.8 W-hr over a period of 8 hrs.

(a)  The power delivered to the headlight is therefore (460.8 W-hrs)

(b)  The current through the headlight is equal to the power it absorbs from the battery divided by the Voltage, at which the power is supplied,

OR                 I = (57.6 W)/(12V) = 4.8 A

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