hw_ch21 - Instructor Tomoyuki Nakayama Tuesday June 1 2010...

Instructor: Tomoyuki Nakayama Tuesday, June 1, 2010
PHY 2005 Applied Physics 2 - Summer C 2010
Solutions for Suggested Homework Problems (Chapter 21)
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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 = I
1
+ I
2
+ I
3.