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6. The current in the circuit is
i
= (150 V – 50 V)/(3.0
Ω
+ 2.0
Ω
) = 20 A.
So from
V
Q
+ 150 V – (2.0
Ω
)
i = V
P
, we get
V
Q
= 100 V + (2.0
Ω
)(20 A) –150 V = –10 V.
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View Full Document 10. (a) The work done by the battery relates to the potential energy change:
( )
12.0V
12.0 eV.
qV e
V e
∆= =
=
(b)
P = iV = neV
= (3.40
×
10
18
/s)(1.60
×
10
–19
C)(12.0 V) = 6.53 W.
18. (a) For each wire,
R
wire
=
ρ
L/A
where
A
=
π
r
2
.
Consequently, we have
R
wire
=
(1.69
×
10
−
8
m
Ω⋅
)(0.200 m)/
π
(0.00100 m)
2
= 0.0011
Ω
.
The total resistive load on the battery is therefore
tot
R
= 2
R
wire
+
R
=
2(
0.0011
Ω) +
6.00
Ω = 6.0022
Ω
.
Dividing this into the battery emf gives the current
tot
12.0 V
1.9993 A
6.0022
i
R
ε
==
=
Ω
.
The voltage across the
R
= 6.00
Ω
resistor is therefore
Vi
R
(1.9993 A)(6.00
Ω
) = 11.996 V
≈
12.0 V.
(b) Similarly, we find the voltagedrop across each wire to be
wire
wire
R
(1.9993 A)(0.0011
Ω
) = 2.15 mV.
(c)
P
=
i
2
R
= (1.9993 A)(6.00
Ω
)
2
= 23.98 W
≈
24.0 W.
(d) Similarly, we find the power dissipated in each wire to be 4.30 mW.
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View Full Document (
V
A
–
V
B
) –
V
3
= 78
−
36 = 42 V,
which implies the current is
i
1
= (42 V)/(2.0
Ω
) = 21 A.
By the junction rule, then, the
current in
R
2
= 4.0
Ω
is
i
2
=
i
1
−
i
= 21 A
−
6.0 A = 15 A.
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This note was uploaded on 05/04/2010 for the course PHYS 11b taught by Professor Mobius during the Spring '09 term at Harvard.
 Spring '09
 Mobius
 Current, Magnetism, Energy, Potential Energy, Work

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