1 Knots and meshes analysis 1

The following network is given below.

• Set up the mesh equation for mesh \(M_1\)!
• Set up the knot equation for the knot \(K_0\)!

1.1 Lösung:

• Mesh equations: \begin {align} M_1 : U_1-U_\mathrm {Ri}-U_\mathrm {R1}-U_\mathrm {R2}-U_\mathrm {R3} &=0 \nonumber \end {align}
• Knot equations: \begin {align} K_0 : +I_1+I_2-I_3+I_4-I_5 &=0 \nonumber \end {align}

2 Knots and meshes analysis 1

The following network is given below.

• How many knots, branches and meshes are needed for analysis?
• Draw a complete tree for the given network.

2.1 Lösung:

• The network has:
  

  • Knot: k=6
  • Branch: k-1=5
  • Mesh: 4

• Complete Tree
  

3 Knots and meshes analysis 3

The following network is given below.

• Mark and label all currents and voltages!
• Set up the equations for all nodes!
• Set up the equations for all meshes!

3.1 Lösung:

• Network with currents and voltages:
  

• Knot equations: \begin {align} K_0 : -I_1+I_3-I_5 &=0 \nonumber \\ K_1 : +I_1-I_3+I_4 &=0 \nonumber \\ K_2 : -I_4+I_6+I_9 &=0 \nonumber \\ K_3 : +I_2-I_8-I_9 &=0 \nonumber \\ K_4 : -I_2-I_7+I_8 &=0 \nonumber \\ K_5 : +I_5-I_6-I_7 &=0 \nonumber \end {align}
• Mesh equations \begin {align} M_1 : U_1-U_\mathrm {Ri}-U_\mathrm {R1}-U_\mathrm {R2}-U_\mathrm {R3} &=0 \nonumber \\ M_2 : U_\mathrm {R3}+U_\mathrm {R4}+U_\mathrm {R5}+U_\mathrm {R6} &=0 \nonumber \\ M_3 : -U_\mathrm {R6}+U_\mathrm {R7}-U_2+U_\mathrm {R9} &=0 \nonumber \\ M_4 : U_2-U_\mathrm {R8} &=0 \nonumber \end {align}

4 Superposition 1

The following network is given below.

Using the superposition method, determine the voltage drop across the resistor \(R_L\). To do this, follow these steps:

• Determine the proportion of \(U_g\).
• Determine the proportion of \(I_g\).
• Combine the results using the superposition theorem.

4.1 Lösung:

• Proportion of \(U_g\):
  

  \begin {equation} U_\mathrm {RL}(U_\mathrm {g}) = \frac {R_\mathrm {L}}{R_\mathrm {i} + R_\mathrm {L}} \cdot U_\mathrm {g} \nonumber \end {equation}

• Proportion of \(I_g\):
  

  \begin {equation} U_\mathrm {RL}(I_\mathrm {g}) = \frac {R_\mathrm {i} \cdot R_\mathrm {i}}{R_\mathrm {i} + R_\mathrm {L}} \cdot I_\mathrm {g} \nonumber \end {equation}

• Superposition principle:
  

  \begin {equation} U_\mathrm {RL} = U_\mathrm {RL}(U_\mathrm {g}) + U_\mathrm {RL}(I_\mathrm {g}) = \frac {R_\mathrm {L}}{R_\mathrm {i} + R_\mathrm {L}} \cdot U_\mathrm {g} + \frac {R_\mathrm {i} \cdot R_\mathrm {i}}{R_\mathrm {i} + R_\mathrm {L}} \cdot I_\mathrm {g} \nonumber \end {equation}

5 Nodal analysis method 1

Given is the network of a real voltage source \(U_\mathrm {q} = 3.6\ V\) with an internal resistance vo \(R_\mathrm {i} = 0,2\ \Omega \).

Convert the specified real voltage source into a real current source. To do this, follow these steps:

• Determine the conductance of the resistance value.
• Convert the real voltage source into a real current source.
• Calculate the current value of the converted current source.

5.1 Lösung:

• Conductance:
  \begin {equation} G_\mathrm {i} = \frac {1}{R_\mathrm {i}} = \frac {1}{0,2\ \Omega } = 5\ \mathrm {S} \nonumber \end {equation}

• Current source:
  

• Current value
  \begin {equation} I_\mathrm {q} = U_\mathrm {q} \cdot G_\mathrm {i} = 3,6\ \mathrm {V} \cdot 5\ \mathrm {S} = 72\ \mathrm {A} \nonumber \end {equation}

6 Nodal analysis method 2

The following network shows two parallel equivalent circuits of battery cells. They consist of voltage sources \(U_1\) and \(U_2\) as well as internal resistances \(R_{i1}\) and \(R_{i2}\). Perform the analysis using the node potential method.

a)

Rearrange the network so that the node potential method can be applied.

b)

Set up the bus admittance matrix.

c)

Set up the vector of the node inflows (I).

d)

Solve the system of equations and calculate the voltage \(U_0\) between terminals \(A\) and \(B\).

6.1 Lösung:

• Conversion:

  \begin {equation} I_1 = \frac {U_1}{R_\mathrm {i1}} = \frac {0,8\ V}{4,8\ \Omega } = 166\ mA \qquad I_2 = \frac {U_2}{R_\mathrm {i2}} = \frac {1,5\ V}{0,8\ \Omega } = 1,875\ A \nonumber \end {equation}

  \begin {equation} G_\mathrm {i1} = \frac {1}{R_\mathrm {i1}} = \frac {1}{4,8\ \Omega } = 208\ m\Omega \qquad G_\mathrm {i2} = \frac {1}{R_\mathrm {i2}} = \frac {1}{0,8\ \Omega } = 1,25\ \Omega \nonumber \end {equation}

• Bus admittance matrix:

  \begin {equation} KAM = (G_\mathrm {i1} + G_\mathrm {i2}) = 208\ m\Omega + 1,25\ \Omega \nonumber \end {equation}

• Vector of node inflows:
  

  \begin {equation} I = (I_1 + I_2) = 166\ mA + 1,875\ A \nonumber \end {equation}

• Solve the system of equations:
  

  \begin {align} KAM \cdot U &= I \nonumber \\ (G_\mathrm {i1} + G_\mathrm {i2}) \cdot U_0 &= (I_1 + I_2) \nonumber \\ (208\ m\Omega + 1,25\ \Omega ) \cdot U_0 &= (166\ mA + 1,875\ A) \nonumber \\ U_0 = \frac {I_1 + I_2}{G_\mathrm {i1} + G_\mathrm {i2}} &= \frac {166\ mA + 1,875\ A}{208\ m\Omega + 1,25\ \Omega } = 1,4\ V \nonumber \end {align}

7 Mesh analysis 1

Given is the network of a real current source \(I_\mathrm {q} = 6\ A\) with a conductance vo \(G_\mathrm {i} = \frac {2}{3}\ S\).

Convert the specified real current source into a real voltage source. To do this, follow these steps:

• Determine the resistance value of the conductance.
• Convert the real current source into a real voltage source.
• Calculate the voltage of the converted voltage source.

7.1 Lösung:

• Resistance value:
  

  \begin {equation} R_\mathrm {i} = \frac {1}{G_\mathrm {i}} = \frac {1}{\frac {2}{3}\ S} = 1,5\ \Omega \nonumber \end {equation}

• Voltage source:
  

• Voltage value
  

  \begin {equation} U_\mathrm {q} = R_\mathrm {i} \cdot I_\mathrm {q} = 1,5\ \mathrm {\Omega } \cdot 6\ \mathrm {A} = 9\ \mathrm {V} \nonumber \end {equation}

8 Mesh analysis 2

The following network shows a network consisting of three resistors and two voltage sources. Perform the analysis of the network using the mesh current method.

Werte: \[ R_1 = 4{,}7\,\text {k}\Omega R_2 = 3{,}3\,\text {k}\Omega R_3 = 2{,}2 \text {k} \Omega U_1 = 24\,\text {V} U_2 = 12\,\text {V} \]

• Preparation of the network
• Define meshes and mesh flows
• Determine resistance matrix
• Assign source voltages
• Set up a system of equations

8.1 Lösung:

• Preparing the network:
  The network consists of:

  • Two Voltage sources: \(U_1\) und \(U_2\),
  • Three resistors: \(R_1\), \(R_2\), \(R_3\),
  • Three possible stitches.

  Voltage sources and resistance values are available.

• Define meshes and mesh flows:
  

  • \(M_1\): Mesh flow in the left mesh,
  • \(M_2\): Mesh flow in the right mesh,
  • The current \(I_\mathrm {R2}\) divided by \(R_2\) results in: \[ I_\mathrm {R2} = \begin {bmatrix} I_{M1} \\ I_{M2} \end {bmatrix} \]
• Determine resistance matrix: \[ \mathbf {R_M} = \begin {bmatrix} R_1 + R_2 & -R_2 \\ -R_2 & R_2 + R_3 \end {bmatrix} \] Inserting the values: \[ \mathbf {R_M} = \begin {bmatrix} 4{,}7\ \Omega + 3{,}3\ \Omega & -3{,}3\ \Omega \\ -3{,}3\ \Omega & 3{,}3\ \Omega + 2{,}2\ \Omega \end {bmatrix} = \begin {bmatrix} 8{,}0\ \Omega & -3{,}3\ \Omega \\ -3{,}3\ \Omega & 5{,}5\ \Omega \end {bmatrix} \]
• Assign source voltages:
  \begin {equation} \mathbf {U_M} = \begin {bmatrix} U_1 \\ -U_2 \end {bmatrix} = \begin {bmatrix} 24\ V \\ -12\ V \end {bmatrix} \nonumber \end {equation}

• Solve a system of equations:
  

  \[ \mathbf {R_M} \cdot \mathbf {I_M} = \mathbf {U_M}, \] whereby: \[ \mathbf {R_M} = \begin {bmatrix} 8{,}0\ \Omega & -3{,}3\ \Omega \\ -3{,}3\ \Omega & 5{,}5\ \Omega \end {bmatrix}, \mathbf {I_M} = \begin {bmatrix} I_{M1} \\ I_{M2} \end {bmatrix}, \mathbf {U_M} = \begin {bmatrix} 24\ V \\ -12\ V \end {bmatrix}. \]

  Solving the system of equations gives: \begin {equation} I_{M1} = 2{,}79\,\text {mA} \text {und} I_{M2} = -0{,}507\,\text {mA} \nonumber \end {equation}

  This results in: \begin {align} I_\mathrm {R1} &= I_\mathrm {M1} = 2,79\ mA \nonumber \\ I_\mathrm {R2} &= I_\mathrm {M1} + (-I_\mathrm {M2}) = 2,79\ mA + 0,507\ mA = 3,297\ mA \nonumber \\ I_\mathrm {R3} &= -I_\mathrm {M2} = 0,507\ mA \nonumber \end {align}

  and

  \begin {align} U_\mathrm {R1} &= R_1 \cdot I_\mathrm {R1} = 4,7\ k\Omega \cdot 2,79\ mA = 13,12\ V \nonumber \\ U_\mathrm {R2} &= R_2 \cdot I_\mathrm {R1} = 3,3\ k\Omega \cdot 3,297\ mA = 10,88\ V \nonumber \\ U_\mathrm {R3} &= R_3 \cdot I_\mathrm {R1} = 2,2\ k\Omega \cdot 0,507\ mA = 1,12\ V \nonumber \end {align}

  Checking the mesh:

  \begin {align} U_\mathrm {1} &= U_\mathrm {R1} + U_\mathrm {R2} = 24\ V \nonumber \\ U_\mathrm {2} &= U_\mathrm {R2} + U_\mathrm {R3} = 12\ V \nonumber \end {align}