For the network of Fig. 8.103: a. Find the currents I 1 and I 2 . b. Determine the voltage V s . Fig. 8.103

Question

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Answer 1

Textbook Question

Chapter 8, Problem 1P

For the network of Fig. 8.103:

a. Find the currents I₁ and I₂.

b. Determine the voltage V_s.

Chapter 8, Problem 1P, For the network of Fig. 8.103: a. Find the currents I1 and I2. b. Determine the voltage Vs. Fig.

Fig. 8.103

Expert Solution

To determine

(a)

Currents $I1$ and $I2$

Answer to Problem 1P

$I1=4.8AI2=1.2A$

Explanation of Solution

Given:

The given electric network is:

Introductory Circuit Analysis; Laboratory Manual For Introductory Circuit Analysis Format: Kit/package/shrinkwrap, Chapter 8, Problem 1P , additional homework tip 1

Calculation:

Use current divider rule:

$I1=6A( R 2 R 1 + R 2 )I1=6A( 8Ω 8Ω+2Ω)I1=4.8A$

Apply KCL:

$I2=6−I1I2=6−4.8I2=1.2A$

Expert Solution

To determine

(b)

Supply voltage $VS$

Answer to Problem 1P

$VS=9.6V$

Explanation of Solution

Given:

The given electric network is:

Introductory Circuit Analysis; Laboratory Manual For Introductory Circuit Analysis Format: Kit/package/shrinkwrap, Chapter 8, Problem 1P , additional homework tip 2

Calculation:

As we can see, the resistors and current source are connected in parallel. Therefore, the voltage across the source is:

$VS=I1R1VS=(4.8A)(2Ω)VS=9.6V$

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Answer 2

Textbook Question

Chapter 8, Problem 1P

For the network of Fig. 8.103:

a. Find the currents I₁ and I₂.

b. Determine the voltage V_s.

Chapter 8, Problem 1P, For the network of Fig. 8.103: a. Find the currents I1 and I2. b. Determine the voltage Vs. Fig.

Fig. 8.103

Expert Solution

To determine

(a)

Currents $I1$ and $I2$

Answer to Problem 1P

$I1=4.8AI2=1.2A$

Explanation of Solution

Given:

The given electric network is:

Introductory Circuit Analysis; Laboratory Manual For Introductory Circuit Analysis Format: Kit/package/shrinkwrap, Chapter 8, Problem 1P , additional homework tip 1

Calculation:

Use current divider rule:

$I1=6A( R 2 R 1 + R 2 )I1=6A( 8Ω 8Ω+2Ω)I1=4.8A$

Apply KCL:

$I2=6−I1I2=6−4.8I2=1.2A$

Expert Solution

To determine

(b)

Supply voltage $VS$

Answer to Problem 1P

$VS=9.6V$

Explanation of Solution

Given:

The given electric network is:

Introductory Circuit Analysis; Laboratory Manual For Introductory Circuit Analysis Format: Kit/package/shrinkwrap, Chapter 8, Problem 1P , additional homework tip 2

Calculation:

As we can see, the resistors and current source are connected in parallel. Therefore, the voltage across the source is:

$VS=I1R1VS=(4.8A)(2Ω)VS=9.6V$

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Answer 3

Textbook Question

Chapter 8, Problem 1P

For the network of Fig. 8.103:

a. Find the currents I₁ and I₂.

b. Determine the voltage V_s.

Chapter 8, Problem 1P, For the network of Fig. 8.103: a. Find the currents I1 and I2. b. Determine the voltage Vs. Fig.

Fig. 8.103

Expert Solution

To determine

(a)

Currents $I1$ and $I2$

Answer to Problem 1P

$I1=4.8AI2=1.2A$

Explanation of Solution

Given:

The given electric network is:

Introductory Circuit Analysis; Laboratory Manual For Introductory Circuit Analysis Format: Kit/package/shrinkwrap, Chapter 8, Problem 1P , additional homework tip 1

Calculation:

Use current divider rule:

$I1=6A( R 2 R 1 + R 2 )I1=6A( 8Ω 8Ω+2Ω)I1=4.8A$

Apply KCL:

$I2=6−I1I2=6−4.8I2=1.2A$

Expert Solution

To determine

(b)

Supply voltage $VS$

Answer to Problem 1P

$VS=9.6V$

Explanation of Solution

Given:

The given electric network is:

Introductory Circuit Analysis; Laboratory Manual For Introductory Circuit Analysis Format: Kit/package/shrinkwrap, Chapter 8, Problem 1P , additional homework tip 2

Calculation:

As we can see, the resistors and current source are connected in parallel. Therefore, the voltage across the source is:

$VS=I1R1VS=(4.8A)(2Ω)VS=9.6V$

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Answer 4

Textbook Question

Chapter 8, Problem 1P

For the network of Fig. 8.103:

a. Find the currents I₁ and I₂.

b. Determine the voltage V_s.

Chapter 8, Problem 1P, For the network of Fig. 8.103: a. Find the currents I1 and I2. b. Determine the voltage Vs. Fig.

Fig. 8.103

Expert Solution

To determine

(a)

Currents $I1$ and $I2$

Answer to Problem 1P

$I1=4.8AI2=1.2A$

Explanation of Solution

Given:

The given electric network is:

Introductory Circuit Analysis; Laboratory Manual For Introductory Circuit Analysis Format: Kit/package/shrinkwrap, Chapter 8, Problem 1P , additional homework tip 1

Calculation:

Use current divider rule:

$I1=6A( R 2 R 1 + R 2 )I1=6A( 8Ω 8Ω+2Ω)I1=4.8A$

Apply KCL:

$I2=6−I1I2=6−4.8I2=1.2A$

Expert Solution

To determine

(b)

Supply voltage $VS$

Answer to Problem 1P

$VS=9.6V$

Explanation of Solution

Given:

The given electric network is:

Introductory Circuit Analysis; Laboratory Manual For Introductory Circuit Analysis Format: Kit/package/shrinkwrap, Chapter 8, Problem 1P , additional homework tip 2

Calculation:

As we can see, the resistors and current source are connected in parallel. Therefore, the voltage across the source is:

$VS=I1R1VS=(4.8A)(2Ω)VS=9.6V$

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

To determine

(a)

Currents $I1$ and $I2$

Answer to Problem 1P

$I1=4.8AI2=1.2A$

Explanation of Solution

Given:

The given electric network is:

Introductory Circuit Analysis; Laboratory Manual For Introductory Circuit Analysis Format: Kit/package/shrinkwrap, Chapter 8, Problem 1P , additional homework tip 1

Calculation:

Use current divider rule:

$I1=6A( R 2 R 1 + R 2 )I1=6A( 8Ω 8Ω+2Ω)I1=4.8A$

Apply KCL:

$I2=6−I1I2=6−4.8I2=1.2A$

Expert Solution

To determine

(b)

Supply voltage $VS$

Answer to Problem 1P

$VS=9.6V$

Explanation of Solution

Given:

The given electric network is:

Introductory Circuit Analysis; Laboratory Manual For Introductory Circuit Analysis Format: Kit/package/shrinkwrap, Chapter 8, Problem 1P , additional homework tip 2

Calculation:

As we can see, the resistors and current source are connected in parallel. Therefore, the voltage across the source is:

$VS=I1R1VS=(4.8A)(2Ω)VS=9.6V$

Answer 8

Expert Solution

To determine

(a)

Currents $I1$ and $I2$

Answer to Problem 1P

$I1=4.8AI2=1.2A$

Explanation of Solution

Given:

The given electric network is:

Introductory Circuit Analysis; Laboratory Manual For Introductory Circuit Analysis Format: Kit/package/shrinkwrap, Chapter 8, Problem 1P , additional homework tip 1

Calculation:

Use current divider rule:

$I1=6A( R 2 R 1 + R 2 )I1=6A( 8Ω 8Ω+2Ω)I1=4.8A$

Apply KCL:

$I2=6−I1I2=6−4.8I2=1.2A$

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a)

Currents $I1$ and $I2$

Answer 13

Answer to Problem 1P

$I1=4.8AI2=1.2A$

Answer 14

Explanation of Solution

Given:

The given electric network is:

Introductory Circuit Analysis; Laboratory Manual For Introductory Circuit Analysis Format: Kit/package/shrinkwrap, Chapter 8, Problem 1P , additional homework tip 1

Calculation:

Use current divider rule:

$I1=6A( R 2 R 1 + R 2 )I1=6A( 8Ω 8Ω+2Ω)I1=4.8A$

Apply KCL:

$I2=6−I1I2=6−4.8I2=1.2A$

Answer 15

Expert Solution

To determine

(b)

Supply voltage $VS$

Answer to Problem 1P

$VS=9.6V$

Explanation of Solution

Given:

The given electric network is:

Introductory Circuit Analysis; Laboratory Manual For Introductory Circuit Analysis Format: Kit/package/shrinkwrap, Chapter 8, Problem 1P , additional homework tip 2

Calculation:

As we can see, the resistors and current source are connected in parallel. Therefore, the voltage across the source is:

$VS=I1R1VS=(4.8A)(2Ω)VS=9.6V$

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

(b)

Supply voltage $VS$

Answer 20

Answer to Problem 1P

$VS=9.6V$

Answer 21

Explanation of Solution

Given:

The given electric network is:

Introductory Circuit Analysis; Laboratory Manual For Introductory Circuit Analysis Format: Kit/package/shrinkwrap, Chapter 8, Problem 1P , additional homework tip 2

Calculation:

As we can see, the resistors and current source are connected in parallel. Therefore, the voltage across the source is:

$VS=I1R1VS=(4.8A)(2Ω)VS=9.6V$