The peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity.

Question

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Answer 1

Question

Chapter 29, Problem 27P

(a)

To determine

To Find: The peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity.

(a)

Expert Solution

Answer to Problem 27P

$ipeak=0.3464 A$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(5 V)cos(ωt+α)$

Where, $α=π6$

Resistance, $R=25 Ω$

Formula used:

Kirchhoff’s Voltage law:

Sum of the potential in a closed loop circuit is zero.

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 29, Problem 27P , additional homework tip 1

ApplyingKirchhoff’s voltage law:

$25i=5cos(ωt−α)+5(ωt+α)25i=(cos(ωt−α)+(ωt+α))5i=2cosωtcosαi=25cosωtcosπ6i=35cosωtipeak=35 Aipeak=0.3464 A$

Conclusion:

$ipeak=0.3464 A$ is the peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity

(b)

To determine

To Find: The peak current in the circuit by using a phasor diagram.

(b)

Expert Solution

Answer to Problem 27P

$ipeak=0.3464 A$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(5 V)cos(ωt+α)$

Where, $α=π6$

Resistance, $R=25 Ω$

Formula used:

Vector formula

$|V→|=|V→1|2+|V→2|2+2|V→1||V→2|cos2α$

Where,

$|V→|$ is the resultant vector

$V→1$ and $V→2$ are the two vectors and $2α$ is the angle between them.

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 29, Problem 27P , additional homework tip 2

$θ1$ and $θ2$ can be given as:

$θ1=ωt−αθ2=ωt+α⇒θ2−θ2=ωt+α−ωt+α⇒θ2−θ2=2α$

Angle between the vectors $V→1$ and $V→2$ is $2α$

Magnitude of $V→$ can be calculated as:

$|V→|=|V→1|2+|V→2|2+2|V→1||V→2|cos2α|V→|=52+52+2×5×5cos2α|V→|=52(1+cos2α)|V→|=52×2cos2α|V→|=10cosα$

But,

$|V→|=IpeakRIpeakR=10cosαIpeak=10cosα25Ipeak=2cosα5$

It is given that

$α=π6$

Therefore,

$Ipeak=25cos(π6)Ipeak=25×32Ipeak=0.3464 A$

Conclusion:

$ipeak=0.3464 A$ is the peak current in the circuit by using a phasor diagram.

(c)

To determine

To Find: The current in the resistor.

(c)

Expert Solution

Answer to Problem 27P

$250[12cosωt−sinωt]$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(7 V)cos(ωt+α)$

Where, $α=π4$

Resistance, $R=25 Ω$

Formula used:

Kirchhoff’s loop rule

Sum of the potential in the circuit is zero.

Calculation:

ApplyingKirchhoff’s loop rule

$iR=V1+V2$

$iR=5cos(ωt−α)+7cos(ωt+α)iR=5cosωtcosα+5sinωtsinα+7cosωtcosα−7sinωtsinαiR=12cosωtcosα−2sinωtsinα$

For,

$α=π4cosπ4=sinπ4=12$

Therefore,

$i×25=12[12cosωt−sinωt]i=250[12cosωt−sinωt]$

Conclusion:

The current in the resistor is $250[12cosωt−sinωt]$ .

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Answer 2

Question

Chapter 29, Problem 27P

(a)

To determine

To Find: The peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity.

(a)

Expert Solution

Answer to Problem 27P

$ipeak=0.3464 A$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(5 V)cos(ωt+α)$

Where, $α=π6$

Resistance, $R=25 Ω$

Formula used:

Kirchhoff’s Voltage law:

Sum of the potential in a closed loop circuit is zero.

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 29, Problem 27P , additional homework tip 1

ApplyingKirchhoff’s voltage law:

$25i=5cos(ωt−α)+5(ωt+α)25i=(cos(ωt−α)+(ωt+α))5i=2cosωtcosαi=25cosωtcosπ6i=35cosωtipeak=35 Aipeak=0.3464 A$

Conclusion:

$ipeak=0.3464 A$ is the peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity

(b)

To determine

To Find: The peak current in the circuit by using a phasor diagram.

(b)

Expert Solution

Answer to Problem 27P

$ipeak=0.3464 A$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(5 V)cos(ωt+α)$

Where, $α=π6$

Resistance, $R=25 Ω$

Formula used:

Vector formula

$|V→|=|V→1|2+|V→2|2+2|V→1||V→2|cos2α$

Where,

$|V→|$ is the resultant vector

$V→1$ and $V→2$ are the two vectors and $2α$ is the angle between them.

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 29, Problem 27P , additional homework tip 2

$θ1$ and $θ2$ can be given as:

$θ1=ωt−αθ2=ωt+α⇒θ2−θ2=ωt+α−ωt+α⇒θ2−θ2=2α$

Angle between the vectors $V→1$ and $V→2$ is $2α$

Magnitude of $V→$ can be calculated as:

$|V→|=|V→1|2+|V→2|2+2|V→1||V→2|cos2α|V→|=52+52+2×5×5cos2α|V→|=52(1+cos2α)|V→|=52×2cos2α|V→|=10cosα$

But,

$|V→|=IpeakRIpeakR=10cosαIpeak=10cosα25Ipeak=2cosα5$

It is given that

$α=π6$

Therefore,

$Ipeak=25cos(π6)Ipeak=25×32Ipeak=0.3464 A$

Conclusion:

$ipeak=0.3464 A$ is the peak current in the circuit by using a phasor diagram.

(c)

To determine

To Find: The current in the resistor.

(c)

Expert Solution

Answer to Problem 27P

$250[12cosωt−sinωt]$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(7 V)cos(ωt+α)$

Where, $α=π4$

Resistance, $R=25 Ω$

Formula used:

Kirchhoff’s loop rule

Sum of the potential in the circuit is zero.

Calculation:

ApplyingKirchhoff’s loop rule

$iR=V1+V2$

$iR=5cos(ωt−α)+7cos(ωt+α)iR=5cosωtcosα+5sinωtsinα+7cosωtcosα−7sinωtsinαiR=12cosωtcosα−2sinωtsinα$

For,

$α=π4cosπ4=sinπ4=12$

Therefore,

$i×25=12[12cosωt−sinωt]i=250[12cosωt−sinωt]$

Conclusion:

The current in the resistor is $250[12cosωt−sinωt]$ .

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Answer 3

Question

Chapter 29, Problem 27P

(a)

To determine

To Find: The peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity.

(a)

Expert Solution

Answer to Problem 27P

$ipeak=0.3464 A$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(5 V)cos(ωt+α)$

Where, $α=π6$

Resistance, $R=25 Ω$

Formula used:

Kirchhoff’s Voltage law:

Sum of the potential in a closed loop circuit is zero.

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 29, Problem 27P , additional homework tip 1

ApplyingKirchhoff’s voltage law:

$25i=5cos(ωt−α)+5(ωt+α)25i=(cos(ωt−α)+(ωt+α))5i=2cosωtcosαi=25cosωtcosπ6i=35cosωtipeak=35 Aipeak=0.3464 A$

Conclusion:

$ipeak=0.3464 A$ is the peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity

(b)

To determine

To Find: The peak current in the circuit by using a phasor diagram.

(b)

Expert Solution

Answer to Problem 27P

$ipeak=0.3464 A$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(5 V)cos(ωt+α)$

Where, $α=π6$

Resistance, $R=25 Ω$

Formula used:

Vector formula

$|V→|=|V→1|2+|V→2|2+2|V→1||V→2|cos2α$

Where,

$|V→|$ is the resultant vector

$V→1$ and $V→2$ are the two vectors and $2α$ is the angle between them.

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 29, Problem 27P , additional homework tip 2

$θ1$ and $θ2$ can be given as:

$θ1=ωt−αθ2=ωt+α⇒θ2−θ2=ωt+α−ωt+α⇒θ2−θ2=2α$

Angle between the vectors $V→1$ and $V→2$ is $2α$

Magnitude of $V→$ can be calculated as:

$|V→|=|V→1|2+|V→2|2+2|V→1||V→2|cos2α|V→|=52+52+2×5×5cos2α|V→|=52(1+cos2α)|V→|=52×2cos2α|V→|=10cosα$

But,

$|V→|=IpeakRIpeakR=10cosαIpeak=10cosα25Ipeak=2cosα5$

It is given that

$α=π6$

Therefore,

$Ipeak=25cos(π6)Ipeak=25×32Ipeak=0.3464 A$

Conclusion:

$ipeak=0.3464 A$ is the peak current in the circuit by using a phasor diagram.

(c)

To determine

To Find: The current in the resistor.

(c)

Expert Solution

Answer to Problem 27P

$250[12cosωt−sinωt]$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(7 V)cos(ωt+α)$

Where, $α=π4$

Resistance, $R=25 Ω$

Formula used:

Kirchhoff’s loop rule

Sum of the potential in the circuit is zero.

Calculation:

ApplyingKirchhoff’s loop rule

$iR=V1+V2$

$iR=5cos(ωt−α)+7cos(ωt+α)iR=5cosωtcosα+5sinωtsinα+7cosωtcosα−7sinωtsinαiR=12cosωtcosα−2sinωtsinα$

For,

$α=π4cosπ4=sinπ4=12$

Therefore,

$i×25=12[12cosωt−sinωt]i=250[12cosωt−sinωt]$

Conclusion:

The current in the resistor is $250[12cosωt−sinωt]$ .

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Answer 4

Question

Chapter 29, Problem 27P

(a)

To determine

To Find: The peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity.

(a)

Expert Solution

Answer to Problem 27P

$ipeak=0.3464 A$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(5 V)cos(ωt+α)$

Where, $α=π6$

Resistance, $R=25 Ω$

Formula used:

Kirchhoff’s Voltage law:

Sum of the potential in a closed loop circuit is zero.

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 29, Problem 27P , additional homework tip 1

ApplyingKirchhoff’s voltage law:

$25i=5cos(ωt−α)+5(ωt+α)25i=(cos(ωt−α)+(ωt+α))5i=2cosωtcosαi=25cosωtcosπ6i=35cosωtipeak=35 Aipeak=0.3464 A$

Conclusion:

$ipeak=0.3464 A$ is the peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity

(b)

To determine

To Find: The peak current in the circuit by using a phasor diagram.

(b)

Expert Solution

Answer to Problem 27P

$ipeak=0.3464 A$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(5 V)cos(ωt+α)$

Where, $α=π6$

Resistance, $R=25 Ω$

Formula used:

Vector formula

$|V→|=|V→1|2+|V→2|2+2|V→1||V→2|cos2α$

Where,

$|V→|$ is the resultant vector

$V→1$ and $V→2$ are the two vectors and $2α$ is the angle between them.

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 29, Problem 27P , additional homework tip 2

$θ1$ and $θ2$ can be given as:

$θ1=ωt−αθ2=ωt+α⇒θ2−θ2=ωt+α−ωt+α⇒θ2−θ2=2α$

Angle between the vectors $V→1$ and $V→2$ is $2α$

Magnitude of $V→$ can be calculated as:

$|V→|=|V→1|2+|V→2|2+2|V→1||V→2|cos2α|V→|=52+52+2×5×5cos2α|V→|=52(1+cos2α)|V→|=52×2cos2α|V→|=10cosα$

But,

$|V→|=IpeakRIpeakR=10cosαIpeak=10cosα25Ipeak=2cosα5$

It is given that

$α=π6$

Therefore,

$Ipeak=25cos(π6)Ipeak=25×32Ipeak=0.3464 A$

Conclusion:

$ipeak=0.3464 A$ is the peak current in the circuit by using a phasor diagram.

(c)

To determine

To Find: The current in the resistor.

(c)

Expert Solution

Answer to Problem 27P

$250[12cosωt−sinωt]$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(7 V)cos(ωt+α)$

Where, $α=π4$

Resistance, $R=25 Ω$

Formula used:

Kirchhoff’s loop rule

Sum of the potential in the circuit is zero.

Calculation:

ApplyingKirchhoff’s loop rule

$iR=V1+V2$

$iR=5cos(ωt−α)+7cos(ωt+α)iR=5cosωtcosα+5sinωtsinα+7cosωtcosα−7sinωtsinαiR=12cosωtcosα−2sinωtsinα$

For,

$α=π4cosπ4=sinπ4=12$

Therefore,

$i×25=12[12cosωt−sinωt]i=250[12cosωt−sinωt]$

Conclusion:

The current in the resistor is $250[12cosωt−sinωt]$ .

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Answer 5

Chapter 29, Problem 27P

(a)

To determine

To Find: The peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity.

(a)

Expert Solution

Answer to Problem 27P

$ipeak=0.3464 A$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(5 V)cos(ωt+α)$

Where, $α=π6$

Resistance, $R=25 Ω$

Formula used:

Kirchhoff’s Voltage law:

Sum of the potential in a closed loop circuit is zero.

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 29, Problem 27P , additional homework tip 1

ApplyingKirchhoff’s voltage law:

$25i=5cos(ωt−α)+5(ωt+α)25i=(cos(ωt−α)+(ωt+α))5i=2cosωtcosαi=25cosωtcosπ6i=35cosωtipeak=35 Aipeak=0.3464 A$

Conclusion:

$ipeak=0.3464 A$ is the peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity

(b)

To determine

To Find: The peak current in the circuit by using a phasor diagram.

(b)

Expert Solution

Answer to Problem 27P

$ipeak=0.3464 A$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(5 V)cos(ωt+α)$

Where, $α=π6$

Resistance, $R=25 Ω$

Formula used:

Vector formula

$|V→|=|V→1|2+|V→2|2+2|V→1||V→2|cos2α$

Where,

$|V→|$ is the resultant vector

$V→1$ and $V→2$ are the two vectors and $2α$ is the angle between them.

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 29, Problem 27P , additional homework tip 2

$θ1$ and $θ2$ can be given as:

$θ1=ωt−αθ2=ωt+α⇒θ2−θ2=ωt+α−ωt+α⇒θ2−θ2=2α$

Angle between the vectors $V→1$ and $V→2$ is $2α$

Magnitude of $V→$ can be calculated as:

$|V→|=|V→1|2+|V→2|2+2|V→1||V→2|cos2α|V→|=52+52+2×5×5cos2α|V→|=52(1+cos2α)|V→|=52×2cos2α|V→|=10cosα$

But,

$|V→|=IpeakRIpeakR=10cosαIpeak=10cosα25Ipeak=2cosα5$

It is given that

$α=π6$

Therefore,

$Ipeak=25cos(π6)Ipeak=25×32Ipeak=0.3464 A$

Conclusion:

$ipeak=0.3464 A$ is the peak current in the circuit by using a phasor diagram.

(c)

To determine

To Find: The current in the resistor.

(c)

Expert Solution

Answer to Problem 27P

$250[12cosωt−sinωt]$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(7 V)cos(ωt+α)$

Where, $α=π4$

Resistance, $R=25 Ω$

Formula used:

Kirchhoff’s loop rule

Sum of the potential in the circuit is zero.

Calculation:

ApplyingKirchhoff’s loop rule

$iR=V1+V2$

$iR=5cos(ωt−α)+7cos(ωt+α)iR=5cosωtcosα+5sinωtsinα+7cosωtcosα−7sinωtsinαiR=12cosωtcosα−2sinωtsinα$

For,

$α=π4cosπ4=sinπ4=12$

Therefore,

$i×25=12[12cosωt−sinωt]i=250[12cosωt−sinωt]$

Conclusion:

The current in the resistor is $250[12cosωt−sinωt]$ .

Answer 6

Chapter 29, Problem 27P

(a)

To determine

To Find: The peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity.

(a)

Expert Solution

Answer to Problem 27P

$ipeak=0.3464 A$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(5 V)cos(ωt+α)$

Where, $α=π6$

Resistance, $R=25 Ω$

Formula used:

Kirchhoff’s Voltage law:

Sum of the potential in a closed loop circuit is zero.

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 29, Problem 27P , additional homework tip 1

ApplyingKirchhoff’s voltage law:

$25i=5cos(ωt−α)+5(ωt+α)25i=(cos(ωt−α)+(ωt+α))5i=2cosωtcosαi=25cosωtcosπ6i=35cosωtipeak=35 Aipeak=0.3464 A$

Conclusion:

$ipeak=0.3464 A$ is the peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity

Answer 7

Chapter 29, Problem 27P

(a)

To determine

To Find: The peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity.

Answer 8

(a)

Expert Solution

Answer to Problem 27P

$ipeak=0.3464 A$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(5 V)cos(ωt+α)$

Where, $α=π6$

Resistance, $R=25 Ω$

Formula used:

Kirchhoff’s Voltage law:

Sum of the potential in a closed loop circuit is zero.

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 29, Problem 27P , additional homework tip 1

ApplyingKirchhoff’s voltage law:

$25i=5cos(ωt−α)+5(ωt+α)25i=(cos(ωt−α)+(ωt+α))5i=2cosωtcosαi=25cosωtcosπ6i=35cosωtipeak=35 Aipeak=0.3464 A$

Conclusion:

$ipeak=0.3464 A$ is the peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity

Answer 13

(b)

To determine

To Find: The peak current in the circuit by using a phasor diagram.

(b)

Expert Solution

Answer to Problem 27P

$ipeak=0.3464 A$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(5 V)cos(ωt+α)$

Where, $α=π6$

Resistance, $R=25 Ω$

Formula used:

Vector formula

$|V→|=|V→1|2+|V→2|2+2|V→1||V→2|cos2α$

Where,

$|V→|$ is the resultant vector

$V→1$ and $V→2$ are the two vectors and $2α$ is the angle between them.

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 29, Problem 27P , additional homework tip 2

$θ1$ and $θ2$ can be given as:

$θ1=ωt−αθ2=ωt+α⇒θ2−θ2=ωt+α−ωt+α⇒θ2−θ2=2α$

Angle between the vectors $V→1$ and $V→2$ is $2α$

Magnitude of $V→$ can be calculated as:

$|V→|=|V→1|2+|V→2|2+2|V→1||V→2|cos2α|V→|=52+52+2×5×5cos2α|V→|=52(1+cos2α)|V→|=52×2cos2α|V→|=10cosα$

But,

$|V→|=IpeakRIpeakR=10cosαIpeak=10cosα25Ipeak=2cosα5$

It is given that

$α=π6$

Therefore,

$Ipeak=25cos(π6)Ipeak=25×32Ipeak=0.3464 A$

Conclusion:

$ipeak=0.3464 A$ is the peak current in the circuit by using a phasor diagram.

Answer 14

(b)

To determine

To Find: The peak current in the circuit by using a phasor diagram.

Answer 15

(b)

Expert Solution

Answer to Problem 27P

$ipeak=0.3464 A$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(5 V)cos(ωt+α)$

Where, $α=π6$

Resistance, $R=25 Ω$

Formula used:

Vector formula

$|V→|=|V→1|2+|V→2|2+2|V→1||V→2|cos2α$

Where,

$|V→|$ is the resultant vector

$V→1$ and $V→2$ are the two vectors and $2α$ is the angle between them.

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 29, Problem 27P , additional homework tip 2

$θ1$ and $θ2$ can be given as:

$θ1=ωt−αθ2=ωt+α⇒θ2−θ2=ωt+α−ωt+α⇒θ2−θ2=2α$

Angle between the vectors $V→1$ and $V→2$ is $2α$

Magnitude of $V→$ can be calculated as:

$|V→|=|V→1|2+|V→2|2+2|V→1||V→2|cos2α|V→|=52+52+2×5×5cos2α|V→|=52(1+cos2α)|V→|=52×2cos2α|V→|=10cosα$

But,

$|V→|=IpeakRIpeakR=10cosαIpeak=10cosα25Ipeak=2cosα5$

It is given that

$α=π6$

Therefore,

$Ipeak=25cos(π6)Ipeak=25×32Ipeak=0.3464 A$

Conclusion:

$ipeak=0.3464 A$ is the peak current in the circuit by using a phasor diagram.

Answer 20

(c)

To determine

To Find: The current in the resistor.

(c)

Expert Solution

Answer to Problem 27P

$250[12cosωt−sinωt]$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(7 V)cos(ωt+α)$

Where, $α=π4$

Resistance, $R=25 Ω$

Formula used:

Kirchhoff’s loop rule

Sum of the potential in the circuit is zero.

Calculation:

ApplyingKirchhoff’s loop rule

$iR=V1+V2$

$iR=5cos(ωt−α)+7cos(ωt+α)iR=5cosωtcosα+5sinωtsinα+7cosωtcosα−7sinωtsinαiR=12cosωtcosα−2sinωtsinα$

For,

$α=π4cosπ4=sinπ4=12$

Therefore,

$i×25=12[12cosωt−sinωt]i=250[12cosωt−sinωt]$

Conclusion:

The current in the resistor is $250[12cosωt−sinωt]$ .

Answer 21

(c)

To determine

To Find: The current in the resistor.

Answer 22

(c)

Expert Solution

Answer to Problem 27P

$250[12cosωt−sinωt]$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V1=(5 V)cos(ωt−α)$

Potential difference across terminals of one of the other generators, $V2=(7 V)cos(ωt+α)$

Where, $α=π4$

Resistance, $R=25 Ω$

Formula used:

Kirchhoff’s loop rule

Sum of the potential in the circuit is zero.

Calculation:

ApplyingKirchhoff’s loop rule

$iR=V1+V2$

$iR=5cos(ωt−α)+7cos(ωt+α)iR=5cosωtcosα+5sinωtsinα+7cosωtcosα−7sinωtsinαiR=12cosωtcosα−2sinωtsinα$

For,

$α=π4cosπ4=sinπ4=12$

Therefore,

$i×25=12[12cosωt−sinωt]i=250[12cosωt−sinωt]$

Conclusion:

The current in the resistor is $250[12cosωt−sinωt]$ .

Answer 27

Ch. 29 - Prob. 1P Ch. 29 - Prob. 2P Ch. 29 - Prob. 3P Ch. 29 - Prob. 4P Ch. 29 - Prob. 5P Ch. 29 - Prob. 6P Ch. 29 - Prob. 7P Ch. 29 - Prob. 8P Ch. 29 - Prob. 9P Ch. 29 - Prob. 10P

The peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity.

Videos

To Find: The peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity.

Answer to Problem 27P

Explanation of Solution

To Find: The peak current in the circuit by using a phasor diagram.

Answer to Problem 27P

Explanation of Solution

To Find: The current in the resistor.

Answer to Problem 27P

Explanation of Solution

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