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Unformatted text preview: case. In circuit A each
branch of the parallel combination consists of two resistances R in series. Thus, the Chapter 20 Problems 1097 resistance of each branch is Req = R + R = 2 R , according to Equation 20.16. The two
parallel branches have an equivalent resistance that can be determined from Equation 20.17
as
1
1
1
=
+
or RA = R
RA 2 R 2 R In circuit B the resistance of circuit A is in parallel with an additional resistance R.
According to Equation 20.17, the equivalent resistance of this combination is
1
1
111
=
+=+
RB RA R R R or RB = 1 R
2 In circuit C the resistance of circuit A is in series with an additional resistance R.
According to Equation 20.16, the equivalent resistance of this combination is
RC = RA + R = R + R = 2 R
We can now use P = V 2 / Req to determine the total power delivered by the battery in each case: [Circuit A ] V 2 ( 6.0 V )
P=
=
= 4.0 W
R
9.0 Ω [Circuit B ] V 2 ( 6.0 V )
P= 1 = 1
= 8.0 W
R 2 ( 9.0 Ω )
2 [Circuit C] V 2 ( 6.0 V )
P=
=
= 2.0 W
2 R 2 ( 9.0 Ω ) 2 2 2 These results are as expected.
______________________________________________________________________________
69. REASONING When two or more resistors are in series, the equivalent resistance is given
by Equation 20.16: Rs = R1 + R2 + R3 + . . . . When resistors are in parallel, the expression to be solved to find the equivalent resistance is Equation 20.17: 1
1
1
1
=+
+
+ ... .
Rp R1 R2 R3 We will use these relations to determine the eight different values of resistance that can be
obtained by connecting together the three resistors: 1.00, 2.00, and 3.00 Ω . 1098 ELECTRIC CIRCUITS SOLUTION When all the resistors are connected in series, the equivalent resistance is the
sum of all three resistors and the equivalent resistance is 6.00 Ω . When all three are in
parallel, we have from Equation 20.17, the equivalent resistance is 0.545 Ω .
We can also connect two of the resistors in parallel and connect the parallel combination in
series with the third resistor. When the 1.00 and 2.00 Ω resistors are connected in parallel
and the 3.00 Ω resistor is connected in series with the parallel combination, the equivalent
resistance is 3.67 Ω . When the 1.00 and 3.00 Ω resistors are connected in parallel and
the 2.00 Ω resistor is connected in series with the parallel combination, the equivalent
resistance is 2.75 Ω . When the 2.00 and 3.00 Ω resistors are connected in parallel and
the 1.00  Ω resistor is connected in series with the parallel combination, the equivalent
resistance is 2.20 Ω . We can also connect two of the resistors in series and put the third resistor in parallel with
the series combination. When the 1.00 and 2.00 Ω resistors are connected in series and the
3.00 Ω resistor is connected in parallel with the series combination, the equivalent
resistance is 1.50 Ω . When the 1.00 and 3.00 Ω resistors are connected in series and
the 2.00 Ω resistor is connected in parallel with the series combination, the equivalent
resistance is 1.33 Ω . Finally, when the 2.00 and 3.00 Ω resistors are connected in series
and the 1.00 Ω resistor is connected in parallel with the series combination, the equivalent
resistance is 0.833 Ω .
______________________________________________________________________________
70. REASONING The power P dissipated in each resistance R is given by Equation 20.6b as
P = I 2 R , where I is the current. This means that we need to determine the current in each
resistor in order to calculate the power. The current in R1 is the same as the current in the
equivalent resistance for the circuit. Since R2 and R3 are in parallel and equal, the current in
R1 splits into two equal parts at the junction A in the circuit.
SOLUTION To determine the equivalent resistance of the circuit, we note that the parallel
combination of R2 and R3 is in series with R1. The equivalent resistance of the parallel
combination can be obtained from Equation 20.17 as follows: 1
1
1
=
+
RP 576 Ω 576 Ω or RP = 288 Ω This 288 resistance is in series with R1, so that the equivalent resistance of the circuit is
given by Equation 20.16 as
Req = 576 Ω + 288 Ω = 864 Ω Chapter 20 Problems 1099 To find the current from the battery we use Ohm’s law:
I= V
120.0 V
=
= 0.139 A
Req
864 Ω Since this is the current in R1, Equation 20.6b gives the power dissipated in R1 as
2
P = I1 R1 = ( 0.139 A ) ( 576 Ω ) = 11.1 W
1
2 R2 and R3 are in parallel and equal, so that the current in R1 splits into two equal parts at the junction A. As a result, there is a current of 1 ( 0.139 A ) in R2 and in R3. Again using
2
Equation 20.6b, we find that the power dissipated in each of these two resistors is
2
P2 = I 2 R2 = 1 ( 0.139 A ) 2 2
P3 = I 3 R3 = 1 ( 0.139 A ) 2 71. 2 ( 576 Ω ) = 2.78 W 2 ( 576 Ω ) = 2.78 W SSM WWW REASONING Since we know that the current in the 8.00Ω resistor is 0.500 A, we can use Ohm's law (V = IR) to find the voltage across the 8.00Ω resistor. The
8.00Ω resistor and the 16.0Ω resistor are in parallel; therefore, the volt...
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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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