EE Pro for TI - 89, 92 Plus
Equations - Resistive Circuits
10
16.1 Resistance Formulas
Four equations in this topic represent the basic relationship
between resistance and conductance. The first equation links the
resistance R of a bar with a length len and a uniform cross-
sectional area A with a resistivity
ρ
. The second equation defines
the conductance G of the same bar in terms of conductivity
σ
, len
and A. The third and fourth equations show the reciprocity of
conductance G resistance R, resistivity
ρ
and conductivity
σ
.
R
len
A
= ⋅
ρ
Eq. 16.1.1
G
A
len
= ⋅
σ
Eq. 16.1.2
G
R
=
1
Eq. 16.1.3
σ
ρ
=
1
Eq. 16.1.4
Example 16.1 -
A copper wire 1500_m long has a resistivity of 6.5_ohm*cm and a cross sectional area of
.45_cm2. Compute the its resistance and conductance.
Solution -
Upon examining the problem, two choices are noted. Equations 16.1.1, 16.1.2 and 16.1.4 or
16.1.1 and 16.1.3 can be used to solve the problem. The second choice was made here. Press
„
to
display the input screen, enter all the known variables and press
„
to solve the selected equation set.
The computed results are shown in the screen display shown here.
Entered Values
Computed results
-PQYP8CTKCDNGUNGP
AO
ρ
AQJOEO
#
AEO
%QORWVGF4GUWNVU4
'AQJO
)
'AUKGOGPU
16.2 Ohm’s Law and Power
The fundamental relationships between voltage, current and power are presented in this section. The first equation
is the classic Ohm's Law, computes the voltage V in terms of the current I, and the resistance R. The next four
equations describe the relationship between power dissipation P, voltage V, current I, resistance R and
conductance G in a variety of alternate forms. The final equation represents the reciprocity between resistance R
and conductance G.