Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 28, Problem 28.16P

Four resistors are connected to a battery as shown in Figure P27.15. (a) Determine the potential difference across each resistor in terms of ε. (b) Determine the current in each resistor in terms of I. (c) What If? If R3 is increased, explain what happens to the current in each of the resistors. (d) In the limit that R3 → ∞, what are the new values of the current in each resistor in terms of I, the original current in the battery?

Figure P27.15

Chapter 28, Problem 28.16P, Four resistors are connected to a battery as shown in Figure P27.15. (a) Determine the potential

(a)

Expert Solution
Check Mark
To determine
The potential difference across each resistor in terms of ε .

Answer to Problem 28.16P

The potential difference across R1 resistor is ε3 , potential difference across R2 resistor is 2ε9 , potential difference across R3 resistor is 4ε9 , potential difference across R4 resistor is 2ε3 .

Explanation of Solution

The resistors R2 and R3 are connected in series combination as shown in figure 1.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 28, Problem 28.16P , additional homework tip  1

Figure (1)

Formula to calculate the resistance across the circuit when resistors R2 and R3 are connected in series combination.

RS=R2+R3 (1)

Here,

RS is the resistance across the circuit when resistors R2 and R3 are connected in series combination.

R2 is the value of resistor 2.

R3 is the value of resistor 3.

Substitute 2R for R2 , 4R for R3 in equation (1) to find RS ,

RS=2R+4R=6R

Thus, the resistance across the circuit when resistors R2 and R3 are connected in series combination is 6R .

The resistors R4 and RS are connected in parallel combination as shown in figure 2.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 28, Problem 28.16P , additional homework tip  2

Figure (2)

Formula to calculate the resistance when the resistors are connected in parallel is,

1RP=1R4+1RS (2)

Here,

RP is the value of the resistance when the resistors are connected in parallel.

R4 is the value of resistor 4.

Substitute 3R for R4 , 6R for RS in equation (2) to find RP ,

1RP=13R+16RRP=2R

Thus, the value of the resistance when the resistors are connected in parallel is 2R .

The resistors R1 and RP are connected in series combination as shown in figure 3.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 28, Problem 28.16P , additional homework tip  3

Figure (3)

Formula to calculate the equivalent resistance across the circuit is,

Req=R1+RP (3)

Here,

Req is the equivalent resistance across the circuit.

R1 is the value of resistor 1.

Substitute R for R1 , 2R for RP in equation (3) to find Req ,

Req=R+2R=3R

Thus, the equivalent resistance across the circuit is 3R .

The equivalent resistance is shown in the figure 4.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 28, Problem 28.16P , additional homework tip  4

Figure (4)

Formula to calculate the current across the circuit is,

I=VReq (4)

Here,

I is the current across the circuit.

V is the voltage across the circuit.

Substitute 3R for Req , ε for V in equation (4) to find I ,

I=ε3R

Thus, the current across the circuit is ε3R .

Formula to calculate the voltage across the RP resistor is,

VRP=IRP (5)

Here,

VRP is the voltage across the RP resistor.

Substitute 2R for RP , ε3R for I in equation (5) to find VRP ,

VRP=(ε3R)2R=2ε3

Thus, the voltage across the RP resistor is 2ε3 .

Thus, the voltage across the R4 resistor and RS resistor is 2ε3 .

Formula to calculate the voltage across the R1 resistor is,

VR1=IR1 (6)

Here,

VR1 is the voltage across the R1 resistor.

Substitute R for R1 , ε3R for I in equation (6) to find VR1 ,

VR1=(ε3R)R=ε3

Thus, the voltage across the R1 resistor is ε3 .

Formula to calculate the current across the RS resistor is,

IRS=VRPRS (7)

Here,

IRS is the current across the RS resistor.

Substitute 6R for RS , 2ε3 for VRP in equation (7) to find IRS ,

IRS=(2ε3)6R=ε9R

Thus, the current across the RS resistor is ε9R .

Formula to calculate the current across the R4 resistor is,

IR4=VRPR4 (8)

Here,

IR4 is the current across the R4 resistor.

Substitute 3R for R4 , 2ε3 for VRP in equation (8) to find IR4 ,

IR4=(2ε3)3R=2ε9R

Thus, the current across the R4 resistor is 2ε9R .

Formula to calculate the voltage across the R2 resistor is,

VR2=IRSR2 (9)

Here,

VR2 is the voltage across the R2 resistor.

Substitute 2R for R2 , ε9R for IRS in equation (9) to find VR2 ,

VR2=(ε9R)2R=2ε9

Thus, the voltage across the R2 resistor is 2ε9 .

Formula to calculate the voltage across the R3 resistor is,

VR3=IRSR3 (10)

Here,

VR3 is the voltage across the R3 resistor.

Substitute 4R for R3 , ε9R for IRS in equation (10) to find VR3 ,

VR3=(ε9R)4R=4ε9

Thus, the voltage across the R3 resistor is 4ε9 .

Conclusion:

Therefore, the potential difference across R1 resistor is ε3 , potential difference across R2 resistor is 2ε9 , potential difference across R3 resistor is 4ε9 , potential difference across R4 resistor is 2ε3 .

(b)

Expert Solution
Check Mark
To determine
The current in each resistor in terms of I .

Answer to Problem 28.16P

The current across R1 resistor is I , current across R2 resistor is I3 , current across R3 resistor is I3 , current across R4 resistor is 2I3 .

Explanation of Solution

Formula to calculate the value of ε is,

V=ε=IReq (11)

Substitute 3R for Req in equation (11) to find ε ,

ε=3IR

Thus, the value of ε is 3IR .

Formula to calculate the current across the R1 resistor is,

IR1=VR1R1 (12)

Here,

IR1 is the current across the R1 resistor.

Substitute R for R1 , ε3 for VR1 in equation (12) to find IR1 ,

IR1=(ε3)R=ε3R (13)

Substitute 3IR for ε in equation (13) to find IR1 ,

IR1=3IR3R=I

Thus, the current across the R1 resistor is I .

Formula to calculate the current across the R2 resistor is,

IR2=VR2R2 (14)

Here,

IR2 is the current across the R2 resistor.

Substitute 2R for R2 , 2ε9 for VR2 in equation (14) to find IR2 ,

IR2=(2ε9)2R=ε9R (15)

Substitute 3IR for ε in equation (15) to find IR2 ,

IR2=3IR9R=I3

Thus, the current across the R2 resistor is I3 .

Formula to calculate the current across the R3 resistor is,

IR3=VR3R3 (16)

Here,

IR3 is the current across the R3 resistor.

Substitute 4R for R3 , 4ε9 for VR3 in equation (16) to find IR3 ,

IR3=(4ε9)4R=ε9R (17)

Substitute 3IR for ε in equation (17) to find IR3 ,

IR3=3IR9R=I3

Thus, the current across the R3 resistor is I3 .

Formula to calculate the current across the R4 resistor is,

IR4=VR4R4 (18)

Here,

IR4 is the current across the R4 resistor.

Substitute 3R for R4 , 2ε3 for VR4 in equation (18) to find IR4 ,

IR4=(2ε3)3R=2ε9R (19)

Substitute 3IR for ε in equation (17) to find IR4 ,

IR4=2(3IR)9R=2I3

Thus, the current across the R4 resistor is 2I3 .

Conclusion:

Therefore, the current across R1 resistor is I , current across R2 resistor is I3 , current across R3 resistor is I3 , current across R4 resistor is 2I3 .

(c)

Expert Solution
Check Mark
To determine
The effect in the current in each of the resistors, if R3 is increased.

Answer to Problem 28.16P

The value of current across R1 resistor increases and the value of current across R1,R2andR4 resistor decreases.

Explanation of Solution

If the value of the R3 is increased, then the value of current across R3 resistor decreases as resistance is inversely proportional to the current.

Since, the current remains same in series combination. So, the value of current across R2 resistor decreases.

If the value of the R3 is increased, then the overall resistance in parallel combination decreases. The voltage remains same in parallel combination. As the voltage is directly proportional to the resistance. Thus, the voltage across the R4 resistor decreases which results in increasing the current across the R4 resistor.

Thus, the current across the R1 resistor decreases as the voltage across the R1 resistor increases.

Conclusion:

Therefore, the value of current across R1 resistor increases and the value of current across R1,R2andR4 resistor decreases.

(d)

Expert Solution
Check Mark
To determine
The new values of the current in each resistor in terms of I and the original current in the battery.

Answer to Problem 28.16P

The current across R1 resistor is 3I4 , current across R2 resistor is 0 , current across R3 resistor is 0 , current across R4 resistor is 3I4 and the original current in the battery is 3I4 .

Explanation of Solution

If R3 tends to infinity then, current across R2 resistor and R3 resistor is 0 .

IR2=IR3=0

Here,

IR2 is the new value of current across R2 resistor.

IR3 is the new value of current across R3 resistor.

The resistors R1 and R4 are connected in series combination.

Formula to calculate the resistance across the circuit when resistors R1 and R4 are connected in series combination.

RS=R1+R4 (20)

Here,

RS is the resistance across the circuit when resistors R1 and R4 are connected in series combination.

R1 is the value of resistor 1.

R4 is the value of resistor 4.

Substitute R for R1 , 3R for R4 in equation (20) to find RS ,

RS=R+3R=4R

Thus, the resistance across the circuit when resistors R1 and R4 are connected in series combination is 4R .

From equation (11), the value of ε is 3IR .

From equation (12), formula to calculate the current across the R1 resistor when R3 tends to infinity is,

IR1=εRS (21)

Here,

IR1 is the new value of current across R1 resistor.

Substitute 3IR for ε , 4R for RS in equation (21) to find IR1 ,

IR1=3IR4R=3I4

Thus, the current across the R1 resistor when R3 tends to infinity is 3I4 .

As the resistors R1 and R4 are connected in series combination so the value of current across R1 resistor is equal to the current across R4 resistor.

IR1=IR4=3I4

Here,

IR4 is the new value of current across R4 resistor.

Thus, the original current in the battery is 3I4 .

Conclusion:

Therefore, the current across R1 resistor is 3I4 , current across R2 resistor is 0 , current across R3 resistor is 0 , current across R4 resistor is 3I4 and the original current in the battery is 3I4 .

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Chapter 28 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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The terminal...Ch. 28 - Two 1.50-V batterieswith their positive terminals...Ch. 28 - An automobile battery has an emf of 12.6 V and 171...Ch. 28 - As in Example 27.2, consider a power supply with...Ch. 28 - Three 100- resistors are connected as shown in...Ch. 28 - Prob. 28.6PCh. 28 - What is the equivalent resistance of the...Ch. 28 - Consider the two circuits shown in Figure P27.5 in...Ch. 28 - Consider the circuit shown in Figure P28.9. 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