Complex circuit solution, Exams of Physics

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COMPLEX CIRCUIT SOLUTION

Draw out the schematic for

the circuit

A

Choose a current orientation for a junction

A

I

I^1

2

I

3

Entering a junction is (+)

Leaving a junction is (-)

According to Kirchhoff’s Current Rule

Equation 1: - I 1

+ I

2

– I

3

nd x

• 1 = MATRIX

Choose “EDIT”, press enter

Change rows to 3 and columns to 4, press enter

Creating a matrix of 3 x 4

Putting the three equations with three unknowns into a matrix

Equation 1: - I 1

+ I

2

– I

3

Equation 2: - 8I 2

– 8I

3

Equation 3: 16I 1

– 8I

3

Enter the coefficient of I 1

, I

2

, and I 3

and the constant in to the matrix

Press enter

All the currents are negative.

This means our original directions

for the current were incorrect.

Revise the direction of junction A.

A

I

I^1

2

I

3

OLD

REVISED

A

I

1

I

2

I

3

I

1

I

2

I

3

Calculating voltage drop across the resistors

V = IR

I

1

= 0.9 A

I

2

= 1.2 A

I

3

= 0.3 A

V = (0.9 A)(8 Ω) = 7.2 V

V = (0.9 A)(8 Ω) = 7.2 V

V = (1.2 A)(8 Ω) = 9.6 V

V = (0.3 A)(8 Ω) = 2.4 V