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Unformatted text preview: ies, so the equivalent resistance R34 of these two is R34 = R3 + R4 = 2R. The resistors R2,
R34, and R5 are in parallel, and the reciprocal of the equivalent resistance R2345 is 1
R2345 = 1
1
1
1
1
15
+
+
=+
+=
R2 R34 R5 R 2 R
R 2R so R2345 = 2R/5. The resistor R1 is in series with R2345, and the equivalent resistance of this
combination is the equivalent resistance of the circuit. Thus, we have R12345 = R1 + R2345 = R + 2R 7R
=
5
5 The power delivered to the circuit is
2 2 V
V
=
R12345 7 R 5
Solving for the resistance R, we find that
P= 5V 2 5 ( 45 V )
R=
=
= 25 Ω
7P
7 ( 58 W )
______________________________________________________________________________
2 118. REASONING AND SOLUTION The resistance of the parallel combination is given by
1
1
1
=+
Rp R 2.00 Ω and the series combination has a resistance of Rs = R + 2.00 Ω. Ohm's law gives for each
case Chapter 20 Problems Ip = VV
V
=+
Rp R 2.00 Ω Is = 1129 V
V
=
Rs R + 2.00 Ω and We know that Ip = 5Is. Using this with the above equations and suppressing units yields the
quadratic equation
R2 – 6.0R + 4.00 = 0
The two solutions for R in this equation are 5.24 Ω and 0.76 Ω .
______________________________________________________________________________
119. SSM REASONING
a. The greatest voltage for the battery is the voltage that generates the maximum current I
that the circuit can tolerate. Once this maximum current is known, the voltage can be
calculated according to Ohm’s law, as the current times the equivalent circuit resistance for
the three resistors in series. To determine the maximum current we note that the power P
dissipated in each resistance R is P = I 2 R according to Equation 20.6b. Since the power
rating and resistance are known for each resistor, the maximum current that can be
tolerated by a resistor is I = P / R . By examining this maximum current for each resistor,
we will be able to identify the maximum current that the circuit can tolerate.
b. The battery delivers power to the circuit that is given by the battery voltage times the
current, according to Equation 20.6a.
SOLUTION
a. Solving Equation 20.6b for the current, we find that the maximum current for each
resistor is as follows: P
4.0 W
I=
=
= 1.4 A
R 42.0 Ω
14444244443
4
2.0Ω resistor 10.0 W
I=
= 0.913 A
12.0 24444
1444 Ω
4
3
12.0Ω resistor 5.0 W
I=
= 1.3 A
3.0 Ω
144 2444
4
3
3.0Ω resistor The smallest of these three values is 0.913 A and is the maximum current that the circuit
can tolerate. Since the resistors are connected in series, the equivalent resistance of the
circuit is
RS = 2.0 Ω + 12.0 Ω + 3.0 Ω = 17.0 Ω
Using Ohm’s law with this equivalent resistance and the maximum current of 0.913 A
reveals that the maximum battery voltage is 1130 ELECTRIC CIRCUITS V = IRS = ( 0.913 A ) (17.0 Ω ) = 15.5 V b. The power delivered by the battery in part (a) is given by Equation 20.6a as
P = IV = ( 0.913 A ) (15.5 V ) = 14.2 W 120. REASONING AND SOLUTION
a. In the first case the parallel resistance of the 75.0 Ω and the 45.0 Ω resistors have an
equivalent resistance that can be calculated using Equation 20.17: 1
1
1
=
+
R p 75.0 Ω 45.0 Ω R p = 28.1 Ω or Ohm’s law, Emf = IR gives Emf = (0.294 A)(28.1 Ω + r), or
Emf = 8.26 V + (0.294 A)r (1) In the second case, Emf = (0.116 A)(75.0 Ω + r), or
(2) Emf = 8.70 V + (0.116 A)r
Multiplying Equation (1) by 0.116 A, Equation (2) by 0.294 A, and subtracting yields
Emf = 8.99 V b. Substituting this result into Equation (1) and solving for r gives r = 2 .5 Ω .
______________________________________________________________________________
121. SSM WWW REASONING We will ignore any changes in length due to thermal
expansion. Although the resistance of each section changes with temperature, the total
resistance of the composite does not change with temperature. Therefore,
R
(144424443 = 144244
) + (R ) R + R 3
tungsten 0 carbon 0 At room temperature tungsten carbon At temperature T From Equation 20.5, we know that the temperature dependence of the resistance for a wire
of resistance R0 at temperature T0 is given by R = R0 [1 + α ( T − T0 )], where α is the
temperature coefficient of resistivity. Thus, (R ) + (R tungsten 0 ) carbon 0 ( = Rtungsten ) 0 (1 + α tungsten ∆T ) + ( Rcarbon )0 (1 + α carbon ∆T ) Chapter 20 Problems 1131 Since ∆ T is the same for each wire, this simplifies to (R )α tungsten 0 tungsten = – ( Rcarbon )0 α carbon (1) This expression can be used to find the ratio of the resistances. Once this ratio is known, we
can find the ratio of the lengths of the sections with the aid of Equation 20.3 (L = RA/ρ).
SOLUTION From Equation (1), the ratio of the resistances of the two sections of the wire
is
–1
Rtungsten
α carbon
–0.0005 [(C°) ] 1
0
=–
=–
=
α tungsten
0.0045 [(C°) –1 ] 9
( Rcarbon ) ( ) 0 Thus, using Equation 20.3, we find the ratio of the tungsten and carbon lengths to be
Ltungsten
Lcarbon ( R0 A / ρ )tungsten ( Rtungsten )0 ρcarbon 1 3.5 × 10–5 Ω ⋅...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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