terminal (current is flowing clockwise), and so
consumes
power. Both bulbs dissipate
energy and so also
consume
power. Note that
P
batt1
= 60 W,
P
bulb1
= 36 W,
P
bulb2
= 4
W, and
P
batt2
= 20 W, so
P
batt1

P
bulb1

P
bulb2

P
batt2
= 0. That’s just conservation of
energy  power delivered minus power dissipated equals zero.
h) The positive charges will be stationary. The free electrons are moving to the
right
,
since the current is pointing to the left.
6.
R
ext
= 0.2
Ω
R
ext
= 11
Ω
a)
I
= 10 Amps
I
= 1 Amp
b)
V
ext
= 2.0 V
V
ext
= 11 V
c)
P
ideal batt
= 120 Watts
P
ideal batt
= 12 Watts
d)
P
ext
= 20 Watts
P
ext
= 11 Watts
e) Energy has been dissipated as heat as a result of current passing through the
internal resistor in both cases. (100 Watts dissipated for the 0.2
Ω
setup, and 1 Watt
dissipated for the 11
Ω
setup) The power delivered by the ideal battery is the same
as the sum of power dissipated by the internal resistor
and
the external resistor.
7. a) i)
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This note was uploaded on 03/10/2012 for the course PHYSICS 8B taught by Professor Shapiro during the Spring '07 term at Berkeley.
 Spring '07
 SHAPIRO

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