(a) The currentthroughR 5 with and without ammeter for the given valuesof the resistor.

Question

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Answer 1

Question

Chapter 2, Problem 2.78HP

To determine

(a)

The currentthroughR₅ with and without ammeter for the given valuesof the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=9.75mA

Without meter in circuit:

IR5=9.91mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 1

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =1 kΩ R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=9.75mA

Without meter in circuit, Rm=0Ω

IR5=9.91mA

To determine

(b)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=22.59mA

Without meter in circuit:

IR5=23.45mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 2

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =100 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=22.59mA

Without meter in circuit, Rm=0Ω

IR5=23.45mA

To determine

(c)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=26.02mA

Without meter in circuit:

IR5=27.16mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 3

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =10 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=26.02mA

Without meter in circuit, Rm=0Ω

IR5=27.16mA

To determine

(d)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=26.42mA

Without meter in circuit:

IR5=27.60mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 4

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =1 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=26.42mA

Without meter in circuit, Rm=0Ω

IR5=27.60mA

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Answer 2

Question

Chapter 2, Problem 2.78HP

To determine

(a)

The currentthroughR₅ with and without ammeter for the given valuesof the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=9.75mA

Without meter in circuit:

IR5=9.91mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 1

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =1 kΩ R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=9.75mA

Without meter in circuit, Rm=0Ω

IR5=9.91mA

To determine

(b)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=22.59mA

Without meter in circuit:

IR5=23.45mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 2

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =100 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=22.59mA

Without meter in circuit, Rm=0Ω

IR5=23.45mA

To determine

(c)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=26.02mA

Without meter in circuit:

IR5=27.16mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 3

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =10 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=26.02mA

Without meter in circuit, Rm=0Ω

IR5=27.16mA

To determine

(d)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=26.42mA

Without meter in circuit:

IR5=27.60mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 4

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =1 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=26.42mA

Without meter in circuit, Rm=0Ω

IR5=27.60mA

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Answer 3

Question

Chapter 2, Problem 2.78HP

To determine

(a)

The currentthroughR₅ with and without ammeter for the given valuesof the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=9.75mA

Without meter in circuit:

IR5=9.91mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 1

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =1 kΩ R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=9.75mA

Without meter in circuit, Rm=0Ω

IR5=9.91mA

To determine

(b)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=22.59mA

Without meter in circuit:

IR5=23.45mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 2

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =100 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=22.59mA

Without meter in circuit, Rm=0Ω

IR5=23.45mA

To determine

(c)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=26.02mA

Without meter in circuit:

IR5=27.16mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 3

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =10 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=26.02mA

Without meter in circuit, Rm=0Ω

IR5=27.16mA

To determine

(d)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=26.42mA

Without meter in circuit:

IR5=27.60mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 4

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =1 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=26.42mA

Without meter in circuit, Rm=0Ω

IR5=27.60mA

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Answer 4

Question

Chapter 2, Problem 2.78HP

To determine

(a)

The currentthroughR₅ with and without ammeter for the given valuesof the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=9.75mA

Without meter in circuit:

IR5=9.91mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 1

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =1 kΩ R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=9.75mA

Without meter in circuit, Rm=0Ω

IR5=9.91mA

To determine

(b)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=22.59mA

Without meter in circuit:

IR5=23.45mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 2

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =100 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=22.59mA

Without meter in circuit, Rm=0Ω

IR5=23.45mA

To determine

(c)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=26.02mA

Without meter in circuit:

IR5=27.16mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 3

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =10 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=26.02mA

Without meter in circuit, Rm=0Ω

IR5=27.16mA

To determine

(d)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=26.42mA

Without meter in circuit:

IR5=27.60mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 4

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =1 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=26.42mA

Without meter in circuit, Rm=0Ω

IR5=27.60mA

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Answer 5

Chapter 2, Problem 2.78HP

To determine

(a)

The currentthroughR₅ with and without ammeter for the given valuesof the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=9.75mA

Without meter in circuit:

IR5=9.91mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 1

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =1 kΩ R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=9.75mA

Without meter in circuit, Rm=0Ω

IR5=9.91mA

To determine

(b)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=22.59mA

Without meter in circuit:

IR5=23.45mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 2

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =100 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=22.59mA

Without meter in circuit, Rm=0Ω

IR5=23.45mA

To determine

(c)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=26.02mA

Without meter in circuit:

IR5=27.16mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 3

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =10 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=26.02mA

Without meter in circuit, Rm=0Ω

IR5=27.16mA

To determine

(d)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=26.42mA

Without meter in circuit:

IR5=27.60mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 4

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =1 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=26.42mA

Without meter in circuit, Rm=0Ω

IR5=27.60mA

Answer 6

Chapter 2, Problem 2.78HP

To determine

(a)

The currentthroughR₅ with and without ammeter for the given valuesof the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=9.75mA

Without meter in circuit:

IR5=9.91mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 1

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =1 kΩ R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=9.75mA

Without meter in circuit, Rm=0Ω

IR5=9.91mA

Answer 7

Chapter 2, Problem 2.78HP

To determine

(a)

The currentthroughR₅ with and without ammeter for the given valuesof the resistor.

Answer 8

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=9.75mA

Without meter in circuit:

IR5=9.91mA

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 1

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =1 kΩ R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=9.75mA

Without meter in circuit, Rm=0Ω

IR5=9.91mA

Answer 13

To determine

(b)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=22.59mA

Without meter in circuit:

IR5=23.45mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 2

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =100 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=22.59mA

Without meter in circuit, Rm=0Ω

IR5=23.45mA

Answer 14

To determine

(b)

The current through R₅ with and without ammeter for the given values of the resistor.

Answer 15

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=22.59mA

Without meter in circuit:

IR5=23.45mA

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 2

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =100 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=22.59mA

Without meter in circuit, Rm=0Ω

IR5=23.45mA

Answer 20

To determine

(c)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=26.02mA

Without meter in circuit:

IR5=27.16mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 3

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =10 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=26.02mA

Without meter in circuit, Rm=0Ω

IR5=27.16mA

Answer 21

To determine

(c)

The current through R₅ with and without ammeter for the given values of the resistor.

Answer 22

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=26.02mA

Without meter in circuit:

IR5=27.16mA

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 3

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =10 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=26.02mA

Without meter in circuit, Rm=0Ω

IR5=27.16mA

Answer 27

To determine

(d)

The current through R₅ with and without ammeter for the given values of the resistor.

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=26.42mA

Without meter in circuit:

IR5=27.60mA

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 4

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =1 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=26.42mA

Without meter in circuit, Rm=0Ω

IR5=27.60mA

Answer 28

To determine

(d)

The current through R₅ with and without ammeter for the given values of the resistor.

Answer 29

Expert Solution

Answer to Problem 2.78HP

The current through resistor:

With meter in circuit:

IR5=26.42mA

Without meter in circuit:

IR5=27.60mA

Answer 30

Expert Solution

Answer 31

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given information:

The given circuit is shown below.

Principles and Applications of Electrical Engineering, Chapter 2, Problem 2.78HP , additional homework tip 4

Also,

R s =20Ω R 1 =800Ω R 2 =600Ω R 3 =1.2 kΩ R 4 =150 Ω R 5 =1 Ω R m =25ΩVs=24 V

Calculation:

First, find an expression for the current through R5 in terms of R5 and the meter resistance, Rm. By the voltage divider rule:

VR3 = R3‖( R 4+ R 5+ R m)⋅VR1R3‖( R 4+ R 5+ R m)+R2

VR1=R1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )⋅VsR1‖( R 3‖ ( R 4 + R 5 + R m )+ R 2 )+Rs

And IR5 = VR3R4+R5+Rm

Therefore,

IR5=R1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )⋅ V sR1‖( R 3 ‖ ( R 4 + R 5 + R m )+ R 2 )+RS⋅R3‖( R 4 + R 5 + R m )R3‖( R 4 + R 5 + R m )+ R 2⋅1R4+R5+Rm= 553.846( R 5 + R m +550)⋅24573.846( R 5 + R m +558.579)⋅1200( R 5 + R m +150)1800( R 5 + R m +550)⋅1( R 5 + R m +150)=15.4424R5+Rm+558.579

Hence for given values of R₅

The current through resistor:

With meter in circuit, Rm=25Ω

IR5=26.42mA

Without meter in circuit, Rm=0Ω

IR5=27.60mA

Answer 34

Ch. 2 - A free electron has an initial potential energy...Ch. 2 - The units for voltage, current, and resistance are...Ch. 2 - A particular fully charged battery can deliver...Ch. 2 - The charge cycle shown in Figure P2.4 is an...Ch. 2 - Batteries (e.g., lead-acid batteries) store...Ch. 2 - What determines: a. The current through an ideal...Ch. 2 - An automotive battery is rated at 120 A-h. This...Ch. 2 - A car battery kept in storage in the basement...Ch. 2 - Suppose the current through a wire is given by the...Ch. 2 - The charge cycle shown in Figure P2.10 is...

(a) The currentthroughR 5 with and without ammeter for the given valuesof the resistor.

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(a)

The currentthroughR₅ with and without ammeter for the given valuesof the resistor.

Answer to Problem 2.78HP

Explanation of Solution

(b)

The current through R₅ with and without ammeter for the given values of the resistor.

Answer to Problem 2.78HP

Explanation of Solution

(c)

The current through R₅ with and without ammeter for the given values of the resistor.

Answer to Problem 2.78HP

Explanation of Solution

(d)

The current through R₅ with and without ammeter for the given values of the resistor.

Answer to Problem 2.78HP

Explanation of Solution

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(a) The currentthroughR 5 with and without ammeter for the given valuesof the resistor.

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