Given the network in Fig. P10.1, (a) write the equations for υ a ( t ) and υ b ( t ) . (b) write the equations for υ c ( t ) and υ d ( t ) .

Question

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Answer 1

Textbook Question

Chapter 10, Problem 1P

Given the network in Fig. P10.1,

(a) write the equations for υ a ( t ) and υ b ( t ) .

(b) write the equations for υ c ( t ) and υ d ( t ) .

Chapter 10, Problem 1P, Given the network in Fig. P10.1, (a) write the equations for a(t)andb(t). (b) write the equations

a

Expert Solution

To determine

The equations for va(t) and vb(t) for the given circuit.

Answer to Problem 1P

The equations are:

va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt

Explanation of Solution

Given:

The given circuit is shown below.

Basic Engineering Circuit Analysis, Chapter 10, Problem 1P , additional homework tip 1

Calculation:

Referring to the given circuit, applying the Kirchhoff’s voltage law to the left side of the loop:

−va(t)+L1di1(t)dt−Mdi2(t)dt=0va(t)=L1di1(t)dt−Mdi2dt ..........(1)

From the circuit:

va(t)=−vc(t) ..........(2)

Solving equation 1 and 2:

va(t)=−vc(t)=L1di1(t)dt−Mdi2dt

Now applying the KVL to the right side of the loop:

−vb(t)−L2di2dt−Mdi1dt=0vb(t)=−L2di2dt−Mdi1dt ............(3)

Therefore, the equation will be:

va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt

b

Expert Solution

To determine

The equations for vc(t) and vd(t) for the given circuit.

Answer to Problem 1P

The equations are:

vc(t)=−L1di1(t)dt+Mdi2dt vd(t)=L2di2dt+Mdi1dt

Explanation of Solution

Given:

The given circuit is shown below.

Basic Engineering Circuit Analysis, Chapter 10, Problem 1P , additional homework tip 2

Calculations:

From part (a), the equations of the va(t) and vb(t) are:

{va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt} ..........(1)

From the circuit:

{va(t)=−vc(t) vb(t)=−vd(t)}...............(2)

Hence, solving the equation:

va(t)=−vc(t)=L1di1(t)dt−Mdi2dt vb(t)=−vd(t)=−L2di2dt−Mdi1dt

Therefore, the equations are:

vc(t)=−L1di1(t)dt+Mdi2dt vd(t)=L2di2dt+Mdi1dt

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Answer 2

Textbook Question

Chapter 10, Problem 1P

Given the network in Fig. P10.1,

(a) write the equations for υ a ( t ) and υ b ( t ) .

(b) write the equations for υ c ( t ) and υ d ( t ) .

Chapter 10, Problem 1P, Given the network in Fig. P10.1, (a) write the equations for a(t)andb(t). (b) write the equations

a

Expert Solution

To determine

The equations for va(t) and vb(t) for the given circuit.

Answer to Problem 1P

The equations are:

va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt

Explanation of Solution

Given:

The given circuit is shown below.

Basic Engineering Circuit Analysis, Chapter 10, Problem 1P , additional homework tip 1

Calculation:

Referring to the given circuit, applying the Kirchhoff’s voltage law to the left side of the loop:

−va(t)+L1di1(t)dt−Mdi2(t)dt=0va(t)=L1di1(t)dt−Mdi2dt ..........(1)

From the circuit:

va(t)=−vc(t) ..........(2)

Solving equation 1 and 2:

va(t)=−vc(t)=L1di1(t)dt−Mdi2dt

Now applying the KVL to the right side of the loop:

−vb(t)−L2di2dt−Mdi1dt=0vb(t)=−L2di2dt−Mdi1dt ............(3)

Therefore, the equation will be:

va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt

b

Expert Solution

To determine

The equations for vc(t) and vd(t) for the given circuit.

Answer to Problem 1P

The equations are:

vc(t)=−L1di1(t)dt+Mdi2dt vd(t)=L2di2dt+Mdi1dt

Explanation of Solution

Given:

The given circuit is shown below.

Basic Engineering Circuit Analysis, Chapter 10, Problem 1P , additional homework tip 2

Calculations:

From part (a), the equations of the va(t) and vb(t) are:

{va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt} ..........(1)

From the circuit:

{va(t)=−vc(t) vb(t)=−vd(t)}...............(2)

Hence, solving the equation:

va(t)=−vc(t)=L1di1(t)dt−Mdi2dt vb(t)=−vd(t)=−L2di2dt−Mdi1dt

Therefore, the equations are:

vc(t)=−L1di1(t)dt+Mdi2dt vd(t)=L2di2dt+Mdi1dt

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Answer 3

Textbook Question

Chapter 10, Problem 1P

Given the network in Fig. P10.1,

(a) write the equations for υ a ( t ) and υ b ( t ) .

(b) write the equations for υ c ( t ) and υ d ( t ) .

Chapter 10, Problem 1P, Given the network in Fig. P10.1, (a) write the equations for a(t)andb(t). (b) write the equations

a

Expert Solution

To determine

The equations for va(t) and vb(t) for the given circuit.

Answer to Problem 1P

The equations are:

va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt

Explanation of Solution

Given:

The given circuit is shown below.

Basic Engineering Circuit Analysis, Chapter 10, Problem 1P , additional homework tip 1

Calculation:

Referring to the given circuit, applying the Kirchhoff’s voltage law to the left side of the loop:

−va(t)+L1di1(t)dt−Mdi2(t)dt=0va(t)=L1di1(t)dt−Mdi2dt ..........(1)

From the circuit:

va(t)=−vc(t) ..........(2)

Solving equation 1 and 2:

va(t)=−vc(t)=L1di1(t)dt−Mdi2dt

Now applying the KVL to the right side of the loop:

−vb(t)−L2di2dt−Mdi1dt=0vb(t)=−L2di2dt−Mdi1dt ............(3)

Therefore, the equation will be:

va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt

b

Expert Solution

To determine

The equations for vc(t) and vd(t) for the given circuit.

Answer to Problem 1P

The equations are:

vc(t)=−L1di1(t)dt+Mdi2dt vd(t)=L2di2dt+Mdi1dt

Explanation of Solution

Given:

The given circuit is shown below.

Basic Engineering Circuit Analysis, Chapter 10, Problem 1P , additional homework tip 2

Calculations:

From part (a), the equations of the va(t) and vb(t) are:

{va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt} ..........(1)

From the circuit:

{va(t)=−vc(t) vb(t)=−vd(t)}...............(2)

Hence, solving the equation:

va(t)=−vc(t)=L1di1(t)dt−Mdi2dt vb(t)=−vd(t)=−L2di2dt−Mdi1dt

Therefore, the equations are:

vc(t)=−L1di1(t)dt+Mdi2dt vd(t)=L2di2dt+Mdi1dt

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Answer 4

Textbook Question

Chapter 10, Problem 1P

Given the network in Fig. P10.1,

(a) write the equations for υ a ( t ) and υ b ( t ) .

(b) write the equations for υ c ( t ) and υ d ( t ) .

Chapter 10, Problem 1P, Given the network in Fig. P10.1, (a) write the equations for a(t)andb(t). (b) write the equations

a

Expert Solution

To determine

The equations for va(t) and vb(t) for the given circuit.

Answer to Problem 1P

The equations are:

va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt

Explanation of Solution

Given:

The given circuit is shown below.

Basic Engineering Circuit Analysis, Chapter 10, Problem 1P , additional homework tip 1

Calculation:

Referring to the given circuit, applying the Kirchhoff’s voltage law to the left side of the loop:

−va(t)+L1di1(t)dt−Mdi2(t)dt=0va(t)=L1di1(t)dt−Mdi2dt ..........(1)

From the circuit:

va(t)=−vc(t) ..........(2)

Solving equation 1 and 2:

va(t)=−vc(t)=L1di1(t)dt−Mdi2dt

Now applying the KVL to the right side of the loop:

−vb(t)−L2di2dt−Mdi1dt=0vb(t)=−L2di2dt−Mdi1dt ............(3)

Therefore, the equation will be:

va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt

b

Expert Solution

To determine

The equations for vc(t) and vd(t) for the given circuit.

Answer to Problem 1P

The equations are:

vc(t)=−L1di1(t)dt+Mdi2dt vd(t)=L2di2dt+Mdi1dt

Explanation of Solution

Given:

The given circuit is shown below.

Basic Engineering Circuit Analysis, Chapter 10, Problem 1P , additional homework tip 2

Calculations:

From part (a), the equations of the va(t) and vb(t) are:

{va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt} ..........(1)

From the circuit:

{va(t)=−vc(t) vb(t)=−vd(t)}...............(2)

Hence, solving the equation:

va(t)=−vc(t)=L1di1(t)dt−Mdi2dt vb(t)=−vd(t)=−L2di2dt−Mdi1dt

Therefore, the equations are:

vc(t)=−L1di1(t)dt+Mdi2dt vd(t)=L2di2dt+Mdi1dt

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a

Expert Solution

To determine

The equations for va(t) and vb(t) for the given circuit.

Answer to Problem 1P

The equations are:

va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt

Explanation of Solution

Given:

The given circuit is shown below.

Basic Engineering Circuit Analysis, Chapter 10, Problem 1P , additional homework tip 1

Calculation:

Referring to the given circuit, applying the Kirchhoff’s voltage law to the left side of the loop:

−va(t)+L1di1(t)dt−Mdi2(t)dt=0va(t)=L1di1(t)dt−Mdi2dt ..........(1)

From the circuit:

va(t)=−vc(t) ..........(2)

Solving equation 1 and 2:

va(t)=−vc(t)=L1di1(t)dt−Mdi2dt

Now applying the KVL to the right side of the loop:

−vb(t)−L2di2dt−Mdi1dt=0vb(t)=−L2di2dt−Mdi1dt ............(3)

Therefore, the equation will be:

va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt

b

Expert Solution

To determine

The equations for vc(t) and vd(t) for the given circuit.

Answer to Problem 1P

The equations are:

vc(t)=−L1di1(t)dt+Mdi2dt vd(t)=L2di2dt+Mdi1dt

Explanation of Solution

Given:

The given circuit is shown below.

Basic Engineering Circuit Analysis, Chapter 10, Problem 1P , additional homework tip 2

Calculations:

From part (a), the equations of the va(t) and vb(t) are:

{va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt} ..........(1)

From the circuit:

{va(t)=−vc(t) vb(t)=−vd(t)}...............(2)

Hence, solving the equation:

va(t)=−vc(t)=L1di1(t)dt−Mdi2dt vb(t)=−vd(t)=−L2di2dt−Mdi1dt

Therefore, the equations are:

vc(t)=−L1di1(t)dt+Mdi2dt vd(t)=L2di2dt+Mdi1dt

Answer 8

a

Expert Solution

To determine

The equations for va(t) and vb(t) for the given circuit.

Answer to Problem 1P

The equations are:

va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt

Explanation of Solution

Given:

The given circuit is shown below.

Basic Engineering Circuit Analysis, Chapter 10, Problem 1P , additional homework tip 1

Calculation:

Referring to the given circuit, applying the Kirchhoff’s voltage law to the left side of the loop:

−va(t)+L1di1(t)dt−Mdi2(t)dt=0va(t)=L1di1(t)dt−Mdi2dt ..........(1)

From the circuit:

va(t)=−vc(t) ..........(2)

Solving equation 1 and 2:

va(t)=−vc(t)=L1di1(t)dt−Mdi2dt

Now applying the KVL to the right side of the loop:

−vb(t)−L2di2dt−Mdi1dt=0vb(t)=−L2di2dt−Mdi1dt ............(3)

Therefore, the equation will be:

va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt

Answer 9

a

Expert Solution

Answer 10

a

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The equations for va(t) and vb(t) for the given circuit.

Answer 13

Answer to Problem 1P

The equations are:

va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt

Answer 14

Explanation of Solution

Given:

The given circuit is shown below.

Basic Engineering Circuit Analysis, Chapter 10, Problem 1P , additional homework tip 1

Calculation:

Referring to the given circuit, applying the Kirchhoff’s voltage law to the left side of the loop:

−va(t)+L1di1(t)dt−Mdi2(t)dt=0va(t)=L1di1(t)dt−Mdi2dt ..........(1)

From the circuit:

va(t)=−vc(t) ..........(2)

Solving equation 1 and 2:

va(t)=−vc(t)=L1di1(t)dt−Mdi2dt

Now applying the KVL to the right side of the loop:

−vb(t)−L2di2dt−Mdi1dt=0vb(t)=−L2di2dt−Mdi1dt ............(3)

Therefore, the equation will be:

va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt

Answer 15

b

Expert Solution

To determine

The equations for vc(t) and vd(t) for the given circuit.

Answer to Problem 1P

The equations are:

vc(t)=−L1di1(t)dt+Mdi2dt vd(t)=L2di2dt+Mdi1dt

Explanation of Solution

Given:

The given circuit is shown below.

Basic Engineering Circuit Analysis, Chapter 10, Problem 1P , additional homework tip 2

Calculations:

From part (a), the equations of the va(t) and vb(t) are:

{va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt} ..........(1)

From the circuit:

{va(t)=−vc(t) vb(t)=−vd(t)}...............(2)

Hence, solving the equation:

va(t)=−vc(t)=L1di1(t)dt−Mdi2dt vb(t)=−vd(t)=−L2di2dt−Mdi1dt

Therefore, the equations are:

vc(t)=−L1di1(t)dt+Mdi2dt vd(t)=L2di2dt+Mdi1dt

Answer 16

b

Expert Solution

Answer 17

b

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The equations for vc(t) and vd(t) for the given circuit.

Answer 20

Answer to Problem 1P

The equations are:

vc(t)=−L1di1(t)dt+Mdi2dt vd(t)=L2di2dt+Mdi1dt

Answer 21

Explanation of Solution

Given:

The given circuit is shown below.

Basic Engineering Circuit Analysis, Chapter 10, Problem 1P , additional homework tip 2

Calculations:

From part (a), the equations of the va(t) and vb(t) are:

{va(t)=L1di1(t)dt−Mdi2dt vb(t)=−L2di2dt−Mdi1dt} ..........(1)

From the circuit:

{va(t)=−vc(t) vb(t)=−vd(t)}...............(2)

Hence, solving the equation:

va(t)=−vc(t)=L1di1(t)dt−Mdi2dt vb(t)=−vd(t)=−L2di2dt−Mdi1dt

Therefore, the equations are:

vc(t)=−L1di1(t)dt+Mdi2dt vd(t)=L2di2dt+Mdi1dt

Answer 22

Ch. 10 - Given the network in Fig. P10.1, (a) write the...

Given the network in Fig. P10.1, (a) write the equations for υ a ( t ) and υ b ( t ) . (b) write the equations for υ c ( t ) and υ d ( t ) .

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Answer to Problem 1P

Explanation of Solution

Answer to Problem 1P

Explanation of Solution

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