This preview shows page 1. Sign up to view the full content.
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)
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '07
 SHAPIRO

Click to edit the document details