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Physics

PHY 303L

Storr

Summer

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rashid (sar3553) – HW6 DC Circuits – storr – (37537) 1
This print-out should have 15 questions. c
Multiple-choice questions may continue on
the next column or page – ﬁnd all choices
18 Ω
before answering.
a 60 Ω
001 10.0 points b
A loop circuit has a resistance of1R and a
current of 1.9 A. The current is reduced to 41 Ω 83 Ω
1.7 A when an additional 3 Ω resistor is added
in series with1R . 96 V
What is the value of 1 ? Assume the inter- d S 1
nal resistance of the source of emf is zero.
Find the resistance between points a and b.
Correct answer: 25.5 Ω.
Correct answer: 12.9024 Ω.
Explanation: Explanation:
c
R1
3
Let : I1= 1.9 A, R
a b
I2= 1.7 A, and
R = 3 Ω.
2 R2 R4
Let the current with R be I and the cur- EB
1 1 d S 1
rent with additional resistanc2 R b2 I . Then
since the emf is a constant and the internal
resistance of the battery is zero, we have
Let : R1= 18 Ω,
R2= 41 Ω,
R3= 60 Ω,
R4= 83 Ω, and
I1R 1 I 2R +1R ) 2
I R = I R + I R E = 96 V.
1 1 2 1 2 2 Ohm’s law is V = I R.
I2
R 1 I − I R2 A good rule of thumb is to eliminate junc-
1 2 tions connected by zero resistance.
= 1.7 A (3 Ω)
1.9 A − 1.7 A R 1
= 25.5 Ω .
a R2 R 3 b
d c
R
002 10.0 points 4
EB
Four resistors are connected as shown in
the ﬁgure. rashid (sar3553) – HW6 DC Circuits – storr – (37537) 2
The series connection of 2 and R 3ives
the equivalent resistance
Let : R 1 65.0 Ω,
R 23= R 2 R 3 R 2 13.0 Ω,
= 41 Ω + 60 Ω R = 13.0 Ω,
3
= 101 Ω. R 4 65.0 Ω, and
Req= 63.0 Ω.
The total resistanceab between a and b can
be obtained by calculating the resistance in R R
the parallel combination of the resistors R , 3 4
1
R 4 and R23 i.e., R R
1 2
1 1 1 1
= + +
R ab R1 R 2 R 3 R4 R E
R4(R 2 R )3+ R R1+ 4 (R 1 R2) 3 S
=
R 1 (4 + 2 ) 3
R1and R 2re in series:
R = R 1 (4 + 2 ) 3
ab R4(R 2 R )3+ R R1+ 4 (R 1 R2) 3 R 12= R 1 R =265 Ω + 13 Ω
= 78 Ω
The denominator is
as are 3 and R4:
R4(R 2 R )3+ R R1+ 4 (R 1 R2) 3
R = R + R = 13 Ω + 65 Ω
= (83 Ω)[41 Ω + 60 Ω] + (18 Ω)(83 Ω) 34 3 4
= 78 Ω .
+ (18 Ω)[41 Ω + 60 Ω]
= 11695 Ω , Then R 12and R 34are in parallel:
−1
so the equivalent resistance is R = 1 + 1
1234 R R
(18 Ω)(83 Ω)[41 Ω + 60 Ω] 12 34
R ab= 2 1 1 −1
(11695 Ω ) = +
78 Ω 78 Ω
= 12.9024 Ω . = 39 Ω
and R is then in series wit1234
003 10.0 points
Req = R + R 1234
The equivalent resistance of the circuit in R = R eq− R1234= 63 Ω − 39 Ω
the ﬁgure is eq= 63.0 Ω.
= 24 Ω .
13 Ω 65 Ω
004 (part 1 of 2) 10.0 points
65 Ω 13 Ω
E
R S
1 Ω 3 V
Find the value of R. . V . 17 Ω .
9 42 17
16.4 Ω 12.1 Ω
Correct answer: 24 Ω.
Explanation: rashid (sar3553) – HW6 DC Circuits – storr – (37537) 3
What is the current through 12.1 Ω bottom- This now gives 2 loop equations and 1 junction
right resistor? equation. The loop equations are
Correct answer: 1.12793 A. V − I R − I R = 0
1 34 34 1 1
Explanation: V2+ I 34R 34− I2R 2 0
c d
b e and the junction equation is
I34+ I2= I 1
E R R E
1 3 4 2
Substituting 34 = I1−I 2n to the loop equa-
R 1 R 2
tions, we have
a g f
h
(R1+ R )34−1R 34 2= V 1
Let : R1= 16.4 Ω, −R 34 1+ (R 2 R )34= 2 . 2
R2= 12.1 Ω,
R = 42.1 Ω, Rewriting and multiplying by factors, we have
3 2
R4= 17 Ω, (R1+ R 34 34 1− R 34I2= R 34 V1
−(R +1 )R34 34 1+(R +1 )(34+R )2 34 2
E1= 9.9 V, and = (R + R )V .
1 34 2
E2= 17.3 V. Subtracting, we have
2
Basic Concepts: Kirchhoﬀ’s Laws [(R1+ R )34 + 2 ) −34 ]I 34 2
X = R 34V1+ (R 1 R )34, 2
V = 0 around a loop.
for
X
I = 0 at a circuit junction.
D = (R 1 R )34 + 2 ) − 34 2
Solution: A key simpliﬁcation is to realize 34
= (16.4 Ω + 12.11 Ω)(12.1 Ω + 12.11 Ω)
that R3and R a4e connected in parallel and 2
can be combined immediately, before applying − (12.11 Ω)
2
Kirchoﬀ’s rules. = 543.575 Ω ,
c d
b e
I I we have
1 2
R 34V1 (R1+ R )34 2
34 34 I2= +
E1 R I E2 D D
(12.11 Ω)(9.9 V)
R 1 R 2 =
543.575 Ω2
a f (16.4 Ω + 12.11 Ω)(17.3 V)
h g +
543.575 Ω2
The combined resistance is given by
= 1.12793 A .
1 1 1 R 3 R 4
R = R + R = R R
34 3 4 3 4
R 3 4 005 (part 2 of 2) 10.0 points
R 34=
R3+ R 4 What is the power dissipated in 17 Ω right-
(42.1 Ω)(17 Ω) centered resistor?
=
42.1 Ω + 17 Ω
= 12.11 Ω. Correct answer: 0.784579 W. rashid (sar3553) – HW6 DC Circuits – storr – (37537) 4
Explanation: E1
In a similar way we can solve for the current r1
through R giving
34 I1
R 2 −1R V 1 2 a r2 b
I34= R R + R (R + R )
1 2 34 1 2 I2
(12.1 Ω)(9.9 V) − (16.4 Ω)(17.3 V)
= E2 r
(16.4 Ω)(12.1 Ω) + (12.11 Ω)(16.4 Ω + 12.1 Ω) 3
= −0.301578 A.
I3
The voltage across R4is just Applying Kirchhoﬀ’s loop rule clockwise
around the right loop,
V 4 V 34 = R 34 34
E1− r 1 1 r I2 20
so the power is
V 2
P = 4 r1 1− E 1 (7 Ω)(2.4 A) − 29 V
R 4 I2= r = (4.8 Ω)
2 2
= (−3.6521 V) = −2.54167 A,
17 Ω
directed from b toward a with magnitude
= 0.784579 W .
2.54167 A .
006 (part 1 of 2) 10.0 points 007 (part 2 of 2) 10.0 points
Find I3.
In th

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