082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

1.2 Laws of indices 3
So
6 2 6 3 =6 5
This illustratesthefirst lawof indiceswhich is
a m a n =a m+n
When expressions with the same base aremultiplied, the indices areadded.
Example1.3 Simplifyeach of the following expressions:
(a) 3 9 3 10 (b) 4 3 4 4 4 6 (c) x 3 x 6 (d) y 4 y 2 y 3
Solution (a) 3 9 3 10 = 3 9+10 = 3 19
(b) 4 3 4 4 4 6 = 4 3+4+6 = 4 13
(c) x 3 x 6 =x 3+6 =x 9
(d) y 4 y 2 y 3 =y 4+2+3 =y 9
Engineeringapplication1.1
Powerdissipationinaresistor
The resistor is one of the three fundamental electronic components. The other two
are the capacitor and the inductor, which we will meet later. The role of the resistor
is to reduce the current flow within the branch of a circuit for a given voltage. As
current flows through the resistor, electrical energy is converted into heat. Because
the energy is lost from the circuit and is effectively wasted, it is termed dissipated
energy. The rate of energy dissipationisknown as the power,P, and isgiven by
P=I 2 R (1.1)
whereI isthecurrentflowingthroughtheresistorandRistheresistancevalue.Note
that the current is raised to the power 2. Note that power, P, is measured in watts;
current,I, ismeasured inamps;and resistance,R, is measured inohms.
Thereisanalternativeformulaforpowerdissipationinaresistorthatusesthevoltage,V,
across the resistor. To obtain this alternative formula we need to use Ohm’s
law,whichstatesthatthevoltageacrossaresistor,V,andthecurrentpassingthrough
it, are relatedby the formula
V=IR (1.2)
From Equation (1.2) wesee that
I = V R
(1.3)
Combining Equations (1.1) and (1.3) gives
( ) V 2
P = R = V ·V V2
·R =
R R R R
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