Please solve all parts of the questions Find the equivalent resistance (in N) between b and c of the 6.00 N and 3.00 N resistors, which are in parallel (right-hand red-brown circles):6.00 QReg, bc3The circuit of equivalent resistances now looks like figure (b). The 6.00Q and 2.00 Q resistors are in series (green circles). Find the equivalent resistance (in N) from a to c:Reg, ac%3DThis resistance is that of the single equivalent resistor in figure (c).(b) What is the current in each resistor if a potential difference of 16.00 V is maintained between a and c?SOLUTIONThe currents in the 2.00 Q and 4.00 N resistors are the same because they are in ---Select---In addition, they carry the same current that would exist in the equivalent resistor subjectthe 16.00 V potential difference.Use the equation R =AV/I and the result from part (a) to find the current (in A) in the 2.00 Q and 4.00 Q resistors:AV ac16.00 VI =Reg,acA%3DReg,acSet the voltages across the resistors in parallel in figure (a) equal to find a relationship between the currents:I2AV, = AV, → (6.00 N)I, = (3.00 N)I2%3D%3D%3DUse I, + I, = I to find I, (in A):%3DI + I2 = I → I, + 2I, = I → I, =1Find I, (in A):I2 = 21, =%3DFinalize As a final check of our results, note that AVpc = (6.00 N)I, = (3.00 N)I, = 4.00 V and AVab = (6.00 N)I = 12.00 V; therefore, the voltage AV (in V) isAV%3D%3D%3D%3D%3DacAVab + AVbc =V, as it must.acEXERCISESuppose that for the figure in the Example, all the resistors are equal to R. The total current in the circuit is 1.90 A, and a 42.0 V battery is connected between a and c. What is R (in 0)?Hint Find the Equivalent ResistanceFour resistors are connected as shown in figure (a), below. (Let R = 2.00 N.)The original network of resistors is reduced to a singleequivalent resistance.6.0 NR4.0ato\ 3.0 N/´Reg, aðReg/ 2.0 NRegaC(a) Find the equivalent resistance between points a and c.SOLUTIONConceptualize Imagine charges flowing into and through this combination from the left. All charges must pass from a to b through the first two resistors, but the charges split at b into---Select-- different paths when encountering the combination of the 6.00 N and the 3.00 N resistors.Ocombinations ofCategorize Because of the simple nature of the combination of resistors in the figure, we categorize this example as one for which we can use the rules for ---Select---resistors.Analyze The combination of resistors can be reduced in steps as shown in the figure.Find the equivalent resistance (in 0) between a and b of the 2.00 N and 4.00 N resistors, which are in series (left-hand red-brown circles):Reg, ab

Imagine charges flowing into and through this combination from the left. All charges must pass from…

Answered: Four resistors are connected as shown…

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Four resistors are connected as shown in figure (a), below. (Let R 2.00 Ω.)

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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Please solve all parts of the questions

Find the equivalent resistance (in N) between b and c of the 6.00 N and 3.00 N resistors, which are in parallel (right-hand red-brown circles):
6.00 Q
Reg, bc
3
The circuit of equivalent resistances now looks like figure (b). The 6.00Q and 2.00 Q resistors are in series (green circles). Find the equivalent resistance (in N) from a to c:
Reg, ac
%3D
This resistance is that of the single equivalent resistor in figure (c).
(b) What is the current in each resistor if a potential difference of 16.00 V is maintained between a and c?
SOLUTION
The currents in the 2.00 Q and 4.00 N resistors are the same because they are in ---Select---
In addition, they carry the same current that would exist in the equivalent resistor subject
the 16.00 V potential difference.
Use the equation R =
AV/I and the result from part (a) to find the current (in A) in the 2.00 Q and 4.00 Q resistors:
AV ac
16.00 V
I =
Reg,ac
A
%3D
Reg,ac
Set the voltages across the resistors in parallel in figure (a) equal to find a relationship between the currents:
I2
AV, = AV, → (6.00 N)I, = (3.00 N)I2
%3D
%3D
%3D
Use I, + I, = I to find I, (in A):
%3D
I + I2 = I → I, + 2I, = I → I, =
1
Find I, (in A):
I2 = 21, =
%3D
Finalize As a final check of our results, note that AVpc = (6.00 N)I, = (3.00 N)I, = 4.00 V and AVab = (6.00 N)I = 12.00 V; therefore, the voltage AV (in V) is
AV
%3D
%3D
%3D
%3D
%3D
ac
AVab + AVbc =
V, as it must.
ac
EXERCISE
Suppose that for the figure in the Example, all the resistors are equal to R. The total current in the circuit is 1.90 A, and a 42.0 V battery is connected between a and c. What is R (in 0)?
Hint

Find the Equivalent Resistance
Four resistors are connected as shown in figure (a), below. (Let R = 2.00 N.)
The original network of resistors is reduced to a single
equivalent resistance.
6.0 N
R
4.0
a
to
\ 3.0 N/
´Reg, að
Reg
/ 2.0 N
Reg
aC
(a) Find the equivalent resistance between points a and c.
SOLUTION
Conceptualize Imagine charges flowing into and through this combination from the left. All charges must pass from a to b through the first two resistors, but the charges split at b into
---Select-- different paths when encountering the combination of the 6.00 N and the 3.00 N resistors.
Ocombinations of
Categorize Because of the simple nature of the combination of resistors in the figure, we categorize this example as one for which we can use the rules for ---Select---
resistors.
Analyze The combination of resistors can be reduced in steps as shown in the figure.
Find the equivalent resistance (in 0) between a and b of the 2.00 N and 4.00 N resistors, which are in series (left-hand red-brown circles):
Reg, ab

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