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Instructor: Tomoyuki Nakayama
Tuesday, June 1, 2010
PHY 2005 Applied Physics 2  Summer C 2010
Solutions for Suggested Homework Problems (Chapter 21)
________________________________________________________________________________
1. (a) Kirchhoff’s loop rule yields 6 + 9 – 45
I
= 0
⇒
I
= 0.333 A. (b) Again we apply the loop rule, we
get 9 6 45
I
= 0
⇒
I
=
0.0667 A. Note that the potential drops if you across a battery from its positive
to negative terminal.
3. (a)
I
2
is flowing into junction A, whereas
I
1
and
I
3
are flowing out. The junction rule gives
I
2
=
I
1
+
I
3.
(b) We the loop rule to obtain +7
I
2
+ 12 +5
I
1
+3
I
1
6 = 0 Note if you traverse a register in the
opposite direction to that of the current flowing through it, the potential rises across the resistor.
7. Two resistors are connected in parallel. The equivalent resistance is
R
eq
= (1/15 + 1/6)
1
=
4.29
Ω
.
The current through the battery is
I
3
= 12/
R
eq
= 2.8 A. The voltage across each resistor is the same as
the voltage provided by the battery, we get
I
1
=
12
/ R
eq
= 0.80 A &
I
2
=
12
/ R
2
= 2.0 A.
10. The current through the 6
Ω
resistor is 0.8/6 = 0.133 A. Applying the loop rule to loop ABDA, we
get 6
I
1
+ 8
I
3
+2 = 0
⇒
I
3
= (6
I
1
– 2)/8 = 0.150 A. The junction rule at point D yields 0 =
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This note was uploaded on 07/16/2011 for the course PHY 2005 taught by Professor Lee during the Summer '08 term at University of Florida.
 Summer '08
 Lee
 Applied Physics, Work

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