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Quiz 5 Solution Guide
Here are the answers and solution guide for quiz 5. Please contact me if you have any
questions.
Jason
[email protected]
Quiz 5 Section A
Version A
Version B
Version C
Version D
1 B
A
D
C
2 A
B
A
C
3 C
B
A
A
4 E
C
A
C
5 D
B
B
E
6 C
D
E
D
Quiz 5 Section B
Version A
Version B
Version C
Version D
1 C
C
B
B
2 B
B
B
D
3 E
A
E
B
4 A
E
A
B
5 B
D
A
A
6 A
B
E
B
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View Full Document Section A Solutions
1.
Solve this expression for drift current,
I
=
nAvQ
where
n
is the density of the
charge carriers,
A
is the cross sectional area of the conductior,
v
is the drift
velocity and
Q
is the charge of each carrier.
2.
Remember, R =
ρ
*L/A. In this instance, A is proportional to r
2
(d
2
) so doubling
both the length and the diameter of the wire will halve the resistance.
3.
Remember the power dissipated by a resistor is P = IV. The battery has internal
resistance (4 ohms), so the voltage drop across the resistor that completes the
circuit will be 36V4.0 A*4ohms = 20V. Hence P = 20V*4A=80 watts.
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This note was uploaded on 04/06/2008 for the course PHYS 2b taught by Professor Schuller during the Spring '08 term at UCSD.
 Spring '08
 schuller
 Physics

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