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NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 20B
RESISTORS IN PARALLEL
PROBLEM
A 42.0 ! resistor is connected in parallel with another resistor across a
9.0 V battery. The current in the circuit is 0.41 A. Calculate the value of
the unknown resistance.
SOLUTION
Given:
∆V = 9.0 V
Unknown:
R2 = ?
R1 = 42.0 Ω
I = 0.41 A
Choose an equation(s) or situation:
Use the equation relating potential difference and equivalent resistance for resistors in parallel, given on page 742.
∆V = IReq
∆V ∆V ∆V
I =  =  + 
Req R1 R2
Rearrange the equation(s) to isolate the unknown(s): Rearrange to solve for R2.
!
"
∆V
∆V
 = I – 
R2
R1
9.0 V
9.0 V
∆V
R2 =  =  =  = 45 Ω
[0.41
A
− 0.21 A]
∆V
9.0 V
I – 
0.41 A − 
R1
42 Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
!
" !
"
ADDITIONAL PRACTICE
1. A 3.3 Ω resistor is connected in parallel with another resistor across a
3.0 V battery. The current in the circuit is 1.41 A. Calculate the value of
the unknown resistance.
2. A 56 Ω resistor is connected in parallel with another resistor across a
12 V battery. The current in the circuit is 3.21 A. Calculate the value of
the unknown resistance.
3. An 18 Ω resistor is connected in parallel with another resistor across a
1.5 V battery. The current in the circuit is 103 mA. Calculate the value
of the unknown resistance.
4. A 39 Ω resistor, an 82 Ω resistor, a 12 Ω resistor and a 22 Ω resistor are
connected in parallel across a potential difference of 3.0 V. Calculate
the equivalent resistance.
5. A 10 Ω resistor, a 12 Ω resistor, a 15 Ω resistor and an 18 Ω resistor are
connected in parallel across a potential difference of 12 V. What is the
equivalent resistance?
Problem 20B
Ch. 20–3
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6. A 33 Ω resistor, a 39 Ω resistor, a 47 Ω resistor and a 68 Ω resistor are
connected in parallel across a potential difference of 1.5 V. Find the
equivalent resistance.
7. A refrigerator and an oven are wired in parallel across a potential difference of 120 V. The refrigerator has a resistance of 75 Ω and the oven has
a resistance of 91 Ω. How much current is in the circuit of each
appliance?
8. A computer and a printer are wired in parallel across a potential difference of 120 V. The computer has a resistance of 82 Ω and the printer
has a resistance of 24 Ω. How much current is in the circuit of each
machine?
9. A lamp and a stereo are wired in parallel across a potential difference of
120 V. The lamp has a resistance of 11 Ω and the stereo has a resistance
of 36 Ω. How much current is in the circuit of each load?
Copyright © by Holt, Rinehart and Winston. All rights reserved.
10. Two bulbs are wired in parallel: one bulb has a resistance of 3.3 Ω, and
the other bulb has a resistance of 4.3 Ω. If the voltage across the circuit
is 1.5 V, what is the current through each bulb?
Ch. 20–4
Holt Physics Problem Bank
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Givens
Solutions
10. R1 = 56 Ω
Req = ΣR = R1 + R2 + R3 = 56Ω + 82Ω + 24Ω = 162Ω
R2 = 82 Ω
R3 = 24 Ω
∆V
9.0 V
I =  =  = 55.6 mA
Req 162 Ω
∆V = 9.0 V
Additional Practice 20B
1. ∆V = 3.0 V
R1 = 3.3 Ω
I = 1.41 A
∆V = IReq
∆V ∆V ∆V
I =  =  + 
Req
R1
R2
!
"
∆V
∆V
 = I − 
R2
R1
3.0 V
∆V
3.0 V
R2 =  =  =  = 6.0 Ω
[1.41 A − 0.91 A]
∆V
3.0 V
I − 
1.41 A − 
R1
3.3 Ω
!
2. ∆V = 12 V
R1 = 56 Ω
I = 3.21 A
" !
"
∆V = IReq
∆V ∆V ∆V
I =  =  + 
Req
R1
R2
!
∆V
∆V
 = I − 
R2
R1
"
12 V
∆V
12 V
R2 =  =  =  = 4.0 Ω
[3.21 A − 0.21 A]
∆V
12 V
I − 
3.21 A − 
R1
56 Ω
!
R1 = 18 Ω
I = 0.103 A
"
∆V = IReq
∆V ∆V ∆V
I =  =  + 
Req
R1
R2
!
"
∆V
∆V
 = I − 
R2
R1
1.5 V
∆V
1.5 V
R2 =  =  =  = 75 Ω
[0.103 A − 0.083 A]
∆V
1.5 V
I − 
0.103 A − 
R1
18 Ω
!
4. R1 = 39 Ω
R2 = 82 Ω
R3 = 12 Ω
R4 = 22 Ω
∆V = 3.0 V
" !
1
1
1
1
1
1
1
1
1
 =  +  +  +  =  +  +  + 
Req R1 R2 R3 R4 39 Ω 82 Ω 12 Ω 22 Ω
1 0.026 0.012 0.083 0.045 0.17
 =  +  +  +  = 
Req 1 Ω
1Ω
1Ω
1Ω
1Ω
Req = 6.0 Ω
V
V Ch. 20–2
"
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. ∆V = 1.5 V
" !
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Givens
5. R1 = 10.0 Ω
R2 = 12 Ω
Print
Solutions
1
1
1
1
1
1
1
1
1
 =  +  +  +  =  +  +  + 
Req R1 R2 R3 R4 10.0 Ω 12 Ω 15 Ω 18 Ω
R3 = 15 Ω
1 0.10 0.083 0.067 0.056
 =  +  +  + 
Req 1 Ω
1Ω
1Ω
1Ω
R4 = 18 Ω
Req = 3.3 Ω
∆V = 12 V
6. R1 = 33 Ω
R2 = 39 Ω
R3 = 47 Ω
R4 = 68 Ω
V = 1.5 V
7. ∆V = 120 V
R1 = 75 Ω
R2 = 91 Ω
1
1
1
1
1
1
1
1
1
 =  +  +  +  =  +  +  + 
Req R1 R2 R3 R4 33 Ω 39 Ω 47 Ω 68 Ω
1
0.030
0.026
0.021
0.015
 =  +  +  + 
Req
1Ω
1Ω
1Ω
1Ω
Req = 11 Ω
V
I1 = 
R1
V
I2 = 
R2
120 V
I1 =  = 1.6 A
75 Ω
120 V
I2 =  = 1.3 A
91 Ω
8. ∆V = 120 V
R1 = 82 Ω
R2 = 24 Ω
V
I1 = 
R1
V
I2 = 
R2
120 V
I1 =  = 1.5 A
82 Ω
Copyright © by Holt, Rinehart and Winston. All rights reserved.
120 V
I2 =  = 5.0 A
24 Ω
9. ∆V = 120 V
R1 = 11 Ω
R2 = 36 Ω
V
I1 = 
R1
V
I2 = 
R2
120 V
I1 =  = 11 A
11 Ω
120 V
I2 =  = 3.3 A
36 Ω
10. ∆V = 1.5 V
R1 = 3.3 Ω
R2 = 4.3 Ω
V
I1 = 
R1
V
I2 = 
R2
1.5 V
I1 =  = 0.45 A
3.3 Ω
1.5 V
I2 =  = 0.35 A
4.3 Ω
V
Section Five—Problem Bank
V Ch. 20–3