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Lecture 08

Branch current method

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Branch current method�

  • A method to analyze a circuit by listing circuit equation with branch currents as unknowns, and then to solve the equations.

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  • Solutions: For a circuit with n nodes and b branches, in order to solve the current of b branches, one can list b independent circuit equations.

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    • (1) Randomly select n-1 nodes from all the n nodes of the circuit to list the KCL equations.

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    • (2) Select b-(n-1) independent loops (Single-link loops or Mesh loops) to list KVL equations.

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Branch current method�

R2

R1

R3

R4

R5

R6

us

+

-

n1

n2

n4

n3

i1

i2

i3

i4

i5

i6

Reference directions: the direction of each branch in the directed graph is the direction of current.

i1 +i2=i6🡪 i1 +i2-i6 = 0 (1)

At node n1

Six branches in the directed graph result in six unknowns (currents).

i3 +i4=i2🡪 i2 –i3-i4 = 0 (2)

At node n2

i4 +i5=i6🡪 i4 +i5-i6 = 0 (3)

At node n3

  • n-1 (n=4) independent KCL equations:
  • b-(n-1), b=6, independent KVL equations:

i2 *R2 + i3*R3 = i1*R1 (4)

i3 *R3 + i5*R5 = i4*R4 (5)

i1 *R1 + i5*R5 + i6*R6 = us (6)

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Branch current method�

  • Procedure of branch current method

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    • Select the associated reference direction of current and voltage of each element (component) and/or voltage/current source (if any) in each branch.

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    • For each of arbitrarily selected n-1 nodes, listing the KCL equations.

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    • Selecting b-(n-1) independent loops and listing the KVL equations for each loop based on VCL.

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  • Note
    • advantage: the branch current method is convenient and intuitive.
    • disadvantage: it can involve quite a lot equations when the circuit graph has many branches.

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Branch current method�

  • Example

7Ω

70V

+

-

6V

+

-

6A

7Ω

+

-

u=?

11Ω

i1

i2

i3

a

b

To solve i1 i2 i3 and the power of each voltage source.

At node a, i1 +i2-i3 = 0 (1)

  • n-1 (n=2) independent KCL equations:
  • b-(n-1) (b=3, n=2) independent KVL equations:

i1 *7 + 6 – i2*11 -70 = 0 (2)

i2 *11 - 6 + i3*7 = 0 (3)

7*i1 – 11*i2 = 64 (2)

i1 + i2 - i3 = 0 (1)

11*i2 + 7*i3 = 6 (3)

1 1 -1

7 -11 0

0 11 7

i1

i2

i3

=

0

64

6

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Branch current method�

1 1 -1

7 -11 0

0 11 7

i1

i2

i3

=

0

64

6

1 1 -1

7 -11 0

0 11 7

i1

i2

i3

=

0

0

0

0

-64

-6

1

1 1 -1

0 -18 7

0 11 7

i1

i2

i3

=

0

0

0

0

-64

-6

1

row2 – row1*7 🡪 row2

1 1 -1

0 -18 7

0 0 203/18

i1

i2

i3

=

0

0

0

0

-64

-406/9

1

row3 + row2*11/18 🡪 row3

i3 = 4A

i2 = -2A

i1 = 6A

P70v = 70*6 = 420W (emitting power)

P6v = -70*2 = -140W (absorbing power)

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Branch current method�

  • Example

7Ω

70V

+

-

6A

7Ω

+

-

11Ω

i1

i2

i3

a

b

+

-

u

  • n-1 (n=2) independent KCL equations:

To solve i1 i2 i3 and the power of each voltage source.

At node a, i1 +i2-i3 = 0 (1)

  • b-(n-1) (b=3, n=2) independent KVL equations:

i1 *7 – i2*11 + u -70 = 0 (2)

i2 *11 + i3*7 – u = 0 (3)

  • We have four unknowns, so one more constraint is needed:

i2 = 6A (4)

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Branch current method�

  • Example

To solve i1 i2 i3 and u.

7Ω

70V

+

-

7Ω

11Ω

i1

i2

i3

a

b

+

-

5u

+

-

u

  • n-1 (n=2) independent KCL equations:

At node a, i1 +i2-i3 = 0 (1)

  • b-(n-1) (b=3, n=2) independent KVL equations:

i1 *7 – i2*11 + 5u -70 = 0 (2)

i2 *11 + i3*7 – 5u = 0 (3)

  • We have four unknowns, so one more constraint is needed:

7*i3 = u (4)

  • Note: when dependent source is present in the circuit, the dependent source is viewed as an independent source in listing the equations.

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