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 28.11P

A battery with ε = 6.00 V and no internal resistance supplies current to the circuit shown in Figure P27.9. When the double-throw switch S is open as shown in the figure, the current in the battery is 1.00 mA. When the switch is closed in position a, the current in the battery is 1.20 mA. When the switch is closed in position b, the current in the battery is 2.00 mA. Find the resistances (a) R1, (b) R2, and (c) R3.

Figure P27.9 Problems 9 and 10.

Chapter 28, Problem 28.11P, A battery with  = 6.00 V and no internal resistance supplies current to the circuit shown in Figure

(a)

Expert Solution
Check Mark
To determine
The value of the resistance R1 .

Answer to Problem 28.11P

The value of the resistance R1 is 1.00kΩ .

Explanation of Solution

Given information: Emf across the battery is 6.00V , current in the battery when the switch S is open is 1.00mA , current in the battery when the switch is closed in position a is 1.20mA , current in the battery when the switch is closed in position b is 2.00mA .

Explanation:

When the switch S is open, then the three resistors R1,R2andR3 are in series.

Formula to calculate the equivalent resistance across the circuit, when the switch S is open.

V=I1Req1Req1=VI1 (1)

Here,

Req1 is the equivalent resistance across the circuit, when the switch S is open.

V is the voltage across the battery.

I1 is the current in the battery when the switch S is open.

As the total emf across the battery is equal to the voltage across the battery.

ε=V

Here,

ε is the total emf across the battery.

Substitute ε for V in equation (1) to find Req1 ,

Req1=εI1 (2)

Formula to calculate the equivalent resistance across the circuit, when the switch S is open.

Req1=R1+R2+R3 (3)

Here,

R1 is the value of resistance 1.

R2 is the value of resistance 2.

R3 is the value of resistance 3.

Substitute εI1 for Req1 in equation (3),

R1+R2+R3=εI1 (4)

Substitute 6.00V for ε , 1.00mA for I1 in equation (4) to find R1+R2+R3 ,

R1+R2+R3=6.00V1.00mA×1A1000mA=6000Ω (5)

When the switch is closed in position a , then the three resistors R1,RPandR3 are in series.

From equation (2), formula to calculate the equivalent resistance across the circuit, when the switch is closed in position a ,

Req2=εI2 (6)

Here,

Req2 is the equivalent resistance across the circuit, when the switch is closed in position a .

I2 is the current in the battery when the switch is closed in position a .

Formula to calculate the resistance when the resistors are connected in parallel.

RP=R2R2R2+R2

From equation (3), formula to calculate the equivalent resistance across the circuit, when the switch is closed in position a .

Req2=R1+RP+R3 (7)

Substitute R2R2R2+R2 for RP in equation (7),

Req2=R1+R2R2R2+R2+R3 (8)

Substitute εI2 for Req2 in equation (8),

R1+R2R2R2+R2+R3=εI2 (9)

Substitute 6.00V for ε , 1.20mA for I2 in equation (9) to find R1+R2R2R2+R2+R3 ,

R1+R2R2R2+R2+R3=6.00V1.20mA×1A1000mAR1+R22+R3=5000Ω (10)

When the switch is closed in position b , then the R3 resistor is short circuited and the two resistors R1andR2 are in series.

From equation (2), formula to calculate the equivalent resistance across the circuit, when the switch is closed in position b ,

Req3=εI3 (11)

Here,

Req3 is the equivalent resistance across the circuit, when the switch is closed in position b .

I3 is the current in the battery when the switch is closed in position a .

From equation (3), formula to calculate the equivalent resistance across the circuit, when the switch is closed in position b .

Req3=R1+R2 (12)

Substitute εI3 for Req3 in equation (12),

R1+R2=εI3 (13)

Substitute 6.00V for ε , 2.00mA for I3 in equation (3) to find R1+R2 ,

R1+R2=6.00V2.00mA×1A1000mA=3000Ω (14)

Subtract equation (14) from (5) to find R3 ,

R1+R2+R3(R1+R2)=6000Ω3000ΩR3=3000Ω=3.00kΩ (15)

Thus, the value of the resistance R3 is 3.00kΩ .

Subtract equation (14) from (10) to find R2 ,

R1+R22+R3(R1+R2)=5000Ω3000ΩR3R22=2000Ω=2.00kΩ (16)

Substitute 3.00kΩ for R3 in equation (16) to find R2 ,

3.00kΩR22=2.00kΩR22=1.00kΩR2=2.00kΩ (17)

Thus, the value of the resistance R2 is 2.00kΩ .

Substitute 3.00kΩ for R3 , 2.00kΩ for R2 in equation (5) to find R1 ,

R1+2.00kΩ+3.00kΩ=6000ΩR1=6000Ω×1kΩ1000Ω2.00kΩ3.00kΩR1=1.00kΩ

Thus, the value of the resistance R1 is 1.00kΩ .

Conclusion:

Therefore, the value of the resistance R1 is 1.00kΩ .

(b)

Expert Solution
Check Mark
To determine
The value of the resistance R2 .

Answer to Problem 28.11P

The value of the resistance R2 is 2.00kΩ .

Explanation of Solution

Given information: Emf across the battery is 6.00V , current in the battery when the switch S is open is 1.00mA , current in the battery when the switch is closed in position a is 1.20mA , current in the battery when the switch is closed in position b is 2.00mA .

Explanation:

From part (a) equation (17), the value of resistance R2 is 2.00kΩ .

Thus, the value of the resistance R2 is 2.00kΩ .

Conclusion:

Therefore, the value of the resistance R2 is 2.00kΩ .

(c)

Expert Solution
Check Mark
To determine
The value of the resistance R3 .

Answer to Problem 28.11P

The value of the resistance R3 is 3.00kΩ .

Explanation of Solution

Given information: Emf across the battery is 6.00V , current in the battery when the switch S is open is 1.00mA , current in the battery when the switch is closed in position a is 1.20mA , current in the battery when the switch is closed in position b is 2.00mA .

Explanation:

From part (a) equation (15), the value of resistance R3 is 3.00kΩ .

Thus, the value of the resistance R3 is 3.00kΩ .

Conclusion:

Therefore, the value of the resistance R3 is 3.00kΩ .

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Chapter 28 Solutions

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

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