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Unformatted text preview: med by the two resistors is 877 W + 576 W = 1450 W .
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56. REASONING
a. The two identical resistors, each with a resistance R, are connected in parallel across the
battery, so the potential difference across each resistor is V, the potential difference provided
V2
by the battery. The power P supplied to each resistor, then, is P =
(Equation 20.6c).
R
The total power supplied by the battery is twice this amount.
b. According to Equation 20.6c, the resistor which is heated until its resistance is 2R
consumes only half as much power as it did initially. Because the resistance of the other
resistor does not change, it consumes the same power as before. The battery supplies less
power, therefore, than it did initially. We will use Equation 20.6c to determine the final
power supplied. 1088 ELECTRIC CIRCUITS SOLUTION
a. Since the power supplied to each resistor is P =
twice as large: Ptot = 2 P = V2
(Equation 20.6c), the total power is
R 2V 2
. Solving for R, we obtain
R
2V 2 2 ( 25 V )
=
= 130 Ω
Ptot
9.6 W
2 R= b. The resistances of the resistors are now R1 = R and R2 = 2R. From Equation 20.6c, the
total power output Ptot = P1 + P2 of the resistors is
V 2 V 2 V 2 V 2 3V 2 3 ( 25 V )
+
=
+
=
=
= 7.2 W
R1 R2
R 2 R 2 R 2 (130 Ω )
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2 P = P + P2 =
1 2
57. REASONING The total power is given by Equation 20.15c as P = Vrms / R p , where Rp is
the equivalent parallel resistance of the heater and the lamp. Since the total power and the
rms voltage are known, we can use this expression to obtain the equivalent parallel
resistance. This equivalent resistance is related to the individual resistances of the heater and
–1
–1
–1
the lamp via Equation 20.17, which is R p = R heater + Rlamp . Since Rheater is given, Rlamp can
be found once Rp is known. SOLUTION According to Equation 20.15c, the equivalent parallel resistance is
Rp = 2
Vrms P Using this result in Equation 20.17 gives
1
1
1
1
=2
=
+
R p Vrms / P Rheater R lamp
Rearranging this expression shows that 1
Rlamp = P
2
Vrms – 1
Rheater = 111 W (120 V ) 2 – 1
4.0 × 10 Ω
2 Therefore, Rlamp = 1
= 190 Ω
5.2 × 10 –3 Ω –1 = 5.2 × 10 –3 Ω –1 Chapter 20 Problems 58. 1089 REASONING The total power P delivered by the battery is related to the equivalent
resistance Req connected between the battery terminals and to the battery voltage V
according to Equation 20.6c: P = V 2 / Req .
When two resistors are connected in series, the equivalent resistance RS of the combination
is greater than the resistance of either resistor alone. This can be seen directly from
RS = R1 + R2 (Equation 20.16).
When two resistors are connected in parallel, the equivalent resistance RP of the
combination is smaller than the resistance of either resistor alone. This can be seen directly
−
−
−
by substituting values in RP 1 = R1 1 + R2 1 (Equation 20.17) or by reviewing the discussion
in Section 20.7 concerning the water flow analogy for electric current in a circuit.
Since the total power delivered by the
battery is P = V 2 / Req and since the power and the battery voltage are the same in both
cases, it follows that the equivalent
resistances are also the same. But the
parallel combination has an equivalent
resistance RP that is smaller than RB,
whereas the series combination has an
equivalent resistance RS that is greater than
RA. This means that RB must be greater
than RA, as Diagram 1 at the right shows. If
RA were greater than RB, as in Diagram 2,
the equivalent resistances RS and RP would
not be equal. Resistance Resistance
RS
RB
RS = RP RB RA
Diagram 1 RA RP
Diagram 2 SOLUTION As discussed in the REASONING, the equivalent resistances in circuits A
and B are equal. According to Equations 20.16 and 20.17, the series and parallel
equivalent resistances are RS = RA + RA = 2 RA
1
1
1
=
+
RP RB RB or Setting the equivalent resistances equal gives RP = 1 RB
2 1090 ELECTRIC CIRCUITS 2 RA = 1 RB
2 RB
=4
RA or As expected, RB is greater than RA. 59. SSM REASONING The equivalent resistance of the three devices in parallel is Rp , and
we can find the value of Rp by using our knowledge of the total power consumption of the
circuit; the value of Rp can be found from Equation 20.6c, P = V 2 / Rp . Ohm's law
(Equation 20.2, V = IR ) can then be used to find the current through the circuit.
SOLUTION
a. The total power used by the circuit is P = 1650 W + 1090 W +1250 W = 3990 W. The
equivalent resistance of the circuit is V 2 (120 V)2
Rp =
=
= 3.6 Ω
P
3990 W
b. The total current through the circuit is
I= V
120 V
=
= 33 A
R p 3.6 Ω This current is larger than the rating of the circuit breaker; therefore, the breaker will open .
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60. REASONING AND SOLUTION The aluminum and copper portions may be viewed a
being connected in parallel since the same voltage appears across them. Using a...
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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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