Find I o in the network in Fig. P5.1 using linearity and the assumption that I o = 1 mA .

Question

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Answer 1

Textbook Question

Chapter 5, Problem 1P

Find I o in the network in Fig. P5.1 using linearity and the assumption that I o = 1 mA .

Chapter 5, Problem 1P, Find Io in the network in Fig. P5.1 using linearity and the assumption that Io=1mA.

Expert Solution & Answer

To determine

The value of current Io using linearity property.

Answer to Problem 1P

The value of Io is 0.5mA.

Explanation of Solution

Given:

A circuit is given with an assumption that Io = 1mA.

Basic Engineering Circuit Analysis, Chapter 5, Problem 1P , additional homework tip 1

Calculations:

Mark the node with voltage V and current in all branches.

Basic Engineering Circuit Analysis, Chapter 5, Problem 1P , additional homework tip 2

As given Io = 1mA passing through the 6kΩ resistor, the voltage across it will be 6V (product of current and resistance).

The value of I3 -

I3=62=3mA

From circuit, I2 is sum of Io and I3

I2 =1+3=4 mA

Applying KVL,

V=3I2+6 =3×4+6V =18V

Therefore, I1 -

I1=V6 =186I1=3mA

Where

IS=I1+I2 =3+4IS =7mA

Let the voltage source be denoted by VS, then apply the KVL in the circuit-

VS=2IS+V =2×7+18 VS =32V

Therefore if

VS=32V→Io=1mAThenVS=64V→Io=2mA

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02:42

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Answer 2

Textbook Question

Chapter 5, Problem 1P

Find I o in the network in Fig. P5.1 using linearity and the assumption that I o = 1 mA .

Chapter 5, Problem 1P, Find Io in the network in Fig. P5.1 using linearity and the assumption that Io=1mA.

Expert Solution & Answer

To determine

The value of current Io using linearity property.

Answer to Problem 1P

The value of Io is 0.5mA.

Explanation of Solution

Given:

A circuit is given with an assumption that Io = 1mA.

Basic Engineering Circuit Analysis, Chapter 5, Problem 1P , additional homework tip 1

Calculations:

Mark the node with voltage V and current in all branches.

Basic Engineering Circuit Analysis, Chapter 5, Problem 1P , additional homework tip 2

As given Io = 1mA passing through the 6kΩ resistor, the voltage across it will be 6V (product of current and resistance).

The value of I3 -

I3=62=3mA

From circuit, I2 is sum of Io and I3

I2 =1+3=4 mA

Applying KVL,

V=3I2+6 =3×4+6V =18V

Therefore, I1 -

I1=V6 =186I1=3mA

Where

IS=I1+I2 =3+4IS =7mA

Let the voltage source be denoted by VS, then apply the KVL in the circuit-

VS=2IS+V =2×7+18 VS =32V

Therefore if

VS=32V→Io=1mAThenVS=64V→Io=2mA

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02:42

Answer 3

Textbook Question

Chapter 5, Problem 1P

Find I o in the network in Fig. P5.1 using linearity and the assumption that I o = 1 mA .

Chapter 5, Problem 1P, Find Io in the network in Fig. P5.1 using linearity and the assumption that Io=1mA.

Expert Solution & Answer

To determine

The value of current Io using linearity property.

Answer to Problem 1P

The value of Io is 0.5mA.

Explanation of Solution

Given:

A circuit is given with an assumption that Io = 1mA.

Basic Engineering Circuit Analysis, Chapter 5, Problem 1P , additional homework tip 1

Calculations:

Mark the node with voltage V and current in all branches.

Basic Engineering Circuit Analysis, Chapter 5, Problem 1P , additional homework tip 2

As given Io = 1mA passing through the 6kΩ resistor, the voltage across it will be 6V (product of current and resistance).

The value of I3 -

I3=62=3mA

From circuit, I2 is sum of Io and I3

I2 =1+3=4 mA

Applying KVL,

V=3I2+6 =3×4+6V =18V

Therefore, I1 -

I1=V6 =186I1=3mA

Where

IS=I1+I2 =3+4IS =7mA

Let the voltage source be denoted by VS, then apply the KVL in the circuit-

VS=2IS+V =2×7+18 VS =32V

Therefore if

VS=32V→Io=1mAThenVS=64V→Io=2mA

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02:42

Answer 4

Textbook Question

Chapter 5, Problem 1P

Find I o in the network in Fig. P5.1 using linearity and the assumption that I o = 1 mA .

Chapter 5, Problem 1P, Find Io in the network in Fig. P5.1 using linearity and the assumption that Io=1mA.

Expert Solution & Answer

To determine

The value of current Io using linearity property.

Answer to Problem 1P

The value of Io is 0.5mA.

Explanation of Solution

Given:

A circuit is given with an assumption that Io = 1mA.

Basic Engineering Circuit Analysis, Chapter 5, Problem 1P , additional homework tip 1

Calculations:

Mark the node with voltage V and current in all branches.

Basic Engineering Circuit Analysis, Chapter 5, Problem 1P , additional homework tip 2

As given Io = 1mA passing through the 6kΩ resistor, the voltage across it will be 6V (product of current and resistance).

The value of I3 -

I3=62=3mA

From circuit, I2 is sum of Io and I3

I2 =1+3=4 mA

Applying KVL,

V=3I2+6 =3×4+6V =18V

Therefore, I1 -

I1=V6 =186I1=3mA

Where

IS=I1+I2 =3+4IS =7mA

Let the voltage source be denoted by VS, then apply the KVL in the circuit-

VS=2IS+V =2×7+18 VS =32V

Therefore if

VS=32V→Io=1mAThenVS=64V→Io=2mA

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02:42

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The value of current Io using linearity property.

Answer to Problem 1P

The value of Io is 0.5mA.

Explanation of Solution

Given:

A circuit is given with an assumption that Io = 1mA.

Basic Engineering Circuit Analysis, Chapter 5, Problem 1P , additional homework tip 1

Calculations:

Mark the node with voltage V and current in all branches.

Basic Engineering Circuit Analysis, Chapter 5, Problem 1P , additional homework tip 2

As given Io = 1mA passing through the 6kΩ resistor, the voltage across it will be 6V (product of current and resistance).

The value of I3 -

I3=62=3mA

From circuit, I2 is sum of Io and I3

I2 =1+3=4 mA

Applying KVL,

V=3I2+6 =3×4+6V =18V

Therefore, I1 -

I1=V6 =186I1=3mA

Where

IS=I1+I2 =3+4IS =7mA

Let the voltage source be denoted by VS, then apply the KVL in the circuit-

VS=2IS+V =2×7+18 VS =32V

Therefore if

VS=32V→Io=1mAThenVS=64V→Io=2mA

Answer 8

Expert Solution & Answer

To determine

The value of current Io using linearity property.

Answer to Problem 1P

The value of Io is 0.5mA.

Explanation of Solution

Given:

A circuit is given with an assumption that Io = 1mA.

Basic Engineering Circuit Analysis, Chapter 5, Problem 1P , additional homework tip 1

Calculations:

Mark the node with voltage V and current in all branches.

Basic Engineering Circuit Analysis, Chapter 5, Problem 1P , additional homework tip 2

As given Io = 1mA passing through the 6kΩ resistor, the voltage across it will be 6V (product of current and resistance).

The value of I3 -

I3=62=3mA

From circuit, I2 is sum of Io and I3

I2 =1+3=4 mA

Applying KVL,

V=3I2+6 =3×4+6V =18V

Therefore, I1 -

I1=V6 =186I1=3mA

Where

IS=I1+I2 =3+4IS =7mA

Let the voltage source be denoted by VS, then apply the KVL in the circuit-

VS=2IS+V =2×7+18 VS =32V

Therefore if

VS=32V→Io=1mAThenVS=64V→Io=2mA

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The value of current Io using linearity property.

Answer 12

Answer to Problem 1P

The value of Io is 0.5mA.

Answer 13

Explanation of Solution

Given:

A circuit is given with an assumption that Io = 1mA.

Basic Engineering Circuit Analysis, Chapter 5, Problem 1P , additional homework tip 1

Calculations:

Mark the node with voltage V and current in all branches.

Basic Engineering Circuit Analysis, Chapter 5, Problem 1P , additional homework tip 2

As given Io = 1mA passing through the 6kΩ resistor, the voltage across it will be 6V (product of current and resistance).

The value of I3 -

I3=62=3mA

From circuit, I2 is sum of Io and I3

I2 =1+3=4 mA

Applying KVL,

V=3I2+6 =3×4+6V =18V

Therefore, I1 -

I1=V6 =186I1=3mA

Where

IS=I1+I2 =3+4IS =7mA

Let the voltage source be denoted by VS, then apply the KVL in the circuit-

VS=2IS+V =2×7+18 VS =32V

Therefore if

VS=32V→Io=1mAThenVS=64V→Io=2mA

Answer 14

Find I o in the network in Fig. P5.1 using linearity and the assumption that I o = 1 mA .

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Answer to Problem 1P

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