The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t 1 . What is the direction of the current I in the top loop (clockwise or counterclockwise) at each of these two times? Figure P6.1 Loops of Problem 6.1.

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 6, Problem 1P

The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t₁. What is the direction of the current I in the top loop (clockwise or counterclockwise) at each of these two times?

Chapter 6, Problem 1P, The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t1.

Figure P6.1 Loops of Problem 6.1.

Expert Solution & Answer

To determine

The direction of the current I in the top loop.

Answer to Problem 1P

The current in the top loop will be in counter-clockwise direction.

Explanation of Solution

Given data:

The required diagram is drawn as shown in Figure 1.

Fundamentals of Applied Electromagnetics (7th Edition), Chapter 6, Problem 1P , additional homework tip 1

Calculation:

The required diagram is drawn as shown in Figure 2 at t=0.

Fundamentals of Applied Electromagnetics (7th Edition), Chapter 6, Problem 1P , additional homework tip 2

It is observed that at t=0, the direction of current in the bottom loop is clockwise. Due to this current, there is an increment of flux in the secondary loop and if flux increases, then the current will also increase.

Hence, the current in the top loop is also momentarily clockwise direction.

From Figure 1, it is observed that there is no current flow in the bottom loop because switch is open. Due to this, there is a decrement of the flux in the secondary loop and if flux decreases then the direction of current will be reversed.

Hence, the current in the top loop will be in counter-clockwise direction.

Conclusion:

Therefore, the current in the top loop will be in counter-clockwise direction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

02:10

Students have asked these similar questions

A dc compound motor having a rating of 10 kW, 1150 r/min, 230 V, 50 A, has the following losses at full-load: bearing friction loss 40 W brush friction loss == 50 W windage loss = 200 W (1) total mechanical losses = 290 W (2) iron losses = 420 W (3) copper loss in the shunt field = 120 W copper losses at full-load: (4) a. in the armature b. in the series field c. in the commutating winding total copper loss in the 500 W 25 W 70 W armature circuit at full-load = 595 W

4 What determines the power rating of a ma- chine? -5 If we cover up the vents in a motor, its out- put power must be reduced. Explain. -6 If a motor operates in a cold environment, may we load it above its rated power? Why?

An electric motor driving a skip hoist with- draws 1.5 metric tons of minerals from a trench 20 m deep every 30 seconds. If the hoist has an overall efficiency of 94 percent, calculate the power output of the motor in horsepower and in kilowatts.

Answer 2

Textbook Question

Chapter 6, Problem 1P

The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t₁. What is the direction of the current I in the top loop (clockwise or counterclockwise) at each of these two times?

Chapter 6, Problem 1P, The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t1.

Figure P6.1 Loops of Problem 6.1.

Expert Solution & Answer

To determine

The direction of the current I in the top loop.

Answer to Problem 1P

The current in the top loop will be in counter-clockwise direction.

Explanation of Solution

Given data:

The required diagram is drawn as shown in Figure 1.

Fundamentals of Applied Electromagnetics (7th Edition), Chapter 6, Problem 1P , additional homework tip 1

Calculation:

The required diagram is drawn as shown in Figure 2 at t=0.

Fundamentals of Applied Electromagnetics (7th Edition), Chapter 6, Problem 1P , additional homework tip 2

It is observed that at t=0, the direction of current in the bottom loop is clockwise. Due to this current, there is an increment of flux in the secondary loop and if flux increases, then the current will also increase.

Hence, the current in the top loop is also momentarily clockwise direction.

From Figure 1, it is observed that there is no current flow in the bottom loop because switch is open. Due to this, there is a decrement of the flux in the secondary loop and if flux decreases then the direction of current will be reversed.

Hence, the current in the top loop will be in counter-clockwise direction.

Conclusion:

Therefore, the current in the top loop will be in counter-clockwise direction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

02:10

Answer 3

Textbook Question

Chapter 6, Problem 1P

The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t₁. What is the direction of the current I in the top loop (clockwise or counterclockwise) at each of these two times?

Chapter 6, Problem 1P, The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t1.

Figure P6.1 Loops of Problem 6.1.

Expert Solution & Answer

To determine

The direction of the current I in the top loop.

Answer to Problem 1P

The current in the top loop will be in counter-clockwise direction.

Explanation of Solution

Given data:

The required diagram is drawn as shown in Figure 1.

Fundamentals of Applied Electromagnetics (7th Edition), Chapter 6, Problem 1P , additional homework tip 1

Calculation:

The required diagram is drawn as shown in Figure 2 at t=0.

Fundamentals of Applied Electromagnetics (7th Edition), Chapter 6, Problem 1P , additional homework tip 2

It is observed that at t=0, the direction of current in the bottom loop is clockwise. Due to this current, there is an increment of flux in the secondary loop and if flux increases, then the current will also increase.

Hence, the current in the top loop is also momentarily clockwise direction.

From Figure 1, it is observed that there is no current flow in the bottom loop because switch is open. Due to this, there is a decrement of the flux in the secondary loop and if flux decreases then the direction of current will be reversed.

Hence, the current in the top loop will be in counter-clockwise direction.

Conclusion:

Therefore, the current in the top loop will be in counter-clockwise direction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

02:10

Answer 4

Textbook Question

Chapter 6, Problem 1P

The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t₁. What is the direction of the current I in the top loop (clockwise or counterclockwise) at each of these two times?

Chapter 6, Problem 1P, The switch in the bottom loop of Fig. P6.1 is closed at t = 0 and then opened at a later time t1.

Figure P6.1 Loops of Problem 6.1.

Expert Solution & Answer

To determine

The direction of the current I in the top loop.

Answer to Problem 1P

The current in the top loop will be in counter-clockwise direction.

Explanation of Solution

Given data:

The required diagram is drawn as shown in Figure 1.

Fundamentals of Applied Electromagnetics (7th Edition), Chapter 6, Problem 1P , additional homework tip 1

Calculation:

The required diagram is drawn as shown in Figure 2 at t=0.

Fundamentals of Applied Electromagnetics (7th Edition), Chapter 6, Problem 1P , additional homework tip 2

It is observed that at t=0, the direction of current in the bottom loop is clockwise. Due to this current, there is an increment of flux in the secondary loop and if flux increases, then the current will also increase.

Hence, the current in the top loop is also momentarily clockwise direction.

From Figure 1, it is observed that there is no current flow in the bottom loop because switch is open. Due to this, there is a decrement of the flux in the secondary loop and if flux decreases then the direction of current will be reversed.

Hence, the current in the top loop will be in counter-clockwise direction.

Conclusion:

Therefore, the current in the top loop will be in counter-clockwise direction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

02:10

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The direction of the current I in the top loop.

Answer to Problem 1P

The current in the top loop will be in counter-clockwise direction.

Explanation of Solution

Given data:

The required diagram is drawn as shown in Figure 1.

Fundamentals of Applied Electromagnetics (7th Edition), Chapter 6, Problem 1P , additional homework tip 1

Calculation:

The required diagram is drawn as shown in Figure 2 at t=0.

Fundamentals of Applied Electromagnetics (7th Edition), Chapter 6, Problem 1P , additional homework tip 2

It is observed that at t=0, the direction of current in the bottom loop is clockwise. Due to this current, there is an increment of flux in the secondary loop and if flux increases, then the current will also increase.

Hence, the current in the top loop is also momentarily clockwise direction.

From Figure 1, it is observed that there is no current flow in the bottom loop because switch is open. Due to this, there is a decrement of the flux in the secondary loop and if flux decreases then the direction of current will be reversed.

Hence, the current in the top loop will be in counter-clockwise direction.

Conclusion:

Therefore, the current in the top loop will be in counter-clockwise direction.

Answer 8

Expert Solution & Answer

To determine

The direction of the current I in the top loop.

Answer to Problem 1P

The current in the top loop will be in counter-clockwise direction.

Explanation of Solution

Given data:

The required diagram is drawn as shown in Figure 1.

Fundamentals of Applied Electromagnetics (7th Edition), Chapter 6, Problem 1P , additional homework tip 1

Calculation:

The required diagram is drawn as shown in Figure 2 at t=0.

Fundamentals of Applied Electromagnetics (7th Edition), Chapter 6, Problem 1P , additional homework tip 2

It is observed that at t=0, the direction of current in the bottom loop is clockwise. Due to this current, there is an increment of flux in the secondary loop and if flux increases, then the current will also increase.

Hence, the current in the top loop is also momentarily clockwise direction.

From Figure 1, it is observed that there is no current flow in the bottom loop because switch is open. Due to this, there is a decrement of the flux in the secondary loop and if flux decreases then the direction of current will be reversed.

Hence, the current in the top loop will be in counter-clockwise direction.

Conclusion:

Therefore, the current in the top loop will be in counter-clockwise direction.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The direction of the current I in the top loop.

Answer 12

Answer to Problem 1P

The current in the top loop will be in counter-clockwise direction.

Answer 13

Explanation of Solution

Given data:

The required diagram is drawn as shown in Figure 1.

Fundamentals of Applied Electromagnetics (7th Edition), Chapter 6, Problem 1P , additional homework tip 1

Calculation:

The required diagram is drawn as shown in Figure 2 at t=0.

Fundamentals of Applied Electromagnetics (7th Edition), Chapter 6, Problem 1P , additional homework tip 2

It is observed that at t=0, the direction of current in the bottom loop is clockwise. Due to this current, there is an increment of flux in the secondary loop and if flux increases, then the current will also increase.

Hence, the current in the top loop is also momentarily clockwise direction.

From Figure 1, it is observed that there is no current flow in the bottom loop because switch is open. Due to this, there is a decrement of the flux in the secondary loop and if flux decreases then the direction of current will be reversed.

Hence, the current in the top loop will be in counter-clockwise direction.

Conclusion:

Therefore, the current in the top loop will be in counter-clockwise direction.