Chapt. 27
Currents and Resistance
211
A to B. Using the two results find the formula for ratio
R
. What does one actually need
to evaluate the formula?
h) Say one had the special case of exact cross sectional symmetry along an axis:
i.e., the
conductor cross section was constant along the axis and the Efield was uniform everywhere
in the conductor and its direction was parallel to the axis of symmetry. In this case, there
is no guide charge. The cross sectional area is
A
and the conductor length is
ℓ
.
027 qfull 00410 2 3 0 moderate math: power integrated for heater
6. A 1250 W radiant heater operates with a potential difference across it of 115 V.
a) What is the current in the heater?
b) What is the resistance
R
0
of the heater coil?
c) How much energy (in MKS units) does the heater absorb and emit (it’s the same amount)
in a time
t
0
= 1 hour?
d) Now say that due to a defect the resistor starts to increase in resistance from its proper
value
R
0
calculated in part (b). The resistance increases according to the expression
R
(
t
) =
R
0
(
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 Fall '06
 Buchler
 Physics, Current, Resistance, Energy, Heat, Light, Incandescent light bulb, incandescent light bulbs

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