Solved Problems

Nodal Analysis

Master Network Analysis

Dr. Mithun Mondal

Demonstrative Video

Problem-1

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Solution-1

  • \[\begin{aligned} \dfrac{V_1-E}{4}+ \dfrac{V_1}{4}+\dfrac{V_1}{2} & =0 \Rightarrow E = 4V_1 \end{aligned}\]
    Applying nodal analysis at node-1
  • \[\begin{aligned} \dfrac{V_2-0}{2}+\dfrac{V_2}{2} & = 4 \Rightarrow V_2 = 4\mathrm{V} \end{aligned}\]
    At Node-2 (last node)
  • \[\begin{aligned} I_R & = \dfrac{0-V_1}{2}+\dfrac{0-V_2}{2} \end{aligned}\]
    is The KCL for
  • \[V_1 = -24~\mathrm{V}\]
    , we get A and Substituting

  • Finally, \(E = 4V_1 = -96~\mathrm{V}\)

  • \(R=0\) means short circuit and the top node voltage is same as ref. node (0V)

Problem-2

  • Using Nodal analysis, find the current in the resistors?

image

Solution-2

  • \[-5+\dfrac{V_1}{1}+\dfrac{V_1-V_2}{1}=0\]
    At Node-1
  • \[-2 + \dfrac{V_2}{2}+\dfrac{V_2-V_1}{1} =0\]
    At Node-2

  • Solving \(V_1= 0.25\) V and \(V_2 = -4.5\) V

  • Current in the resistors can be found using ohm’s law \(I=V/R\)

Problem-3

  • Find the values of \(V1\) and \(V2\) ?

image

Solution-3

  • \[-5+\dfrac{V}{4}+\dfrac{V}{4}+2I=0\]
    Applying KCl at Node
  • \[I = \dfrac{V}{4}\]
    Also
  • \[V_1 = 5~\mathrm{V} \quad V_2 = 25~\mathrm{V}\]
    Solution

Problem-4

  • Find the Node voltages?

image

Solution-4

  • \[v_2-v_1 = 2\]
    Super-node voltage constraint equation:
  • \[v_1 = -7.333~\mathrm{V} \quad v_2 = -5.333~\mathrm{V}\]
    On solving, we get

  • Note: 10\(\Omega\) resistor do not make any difference because it is connected across super-node.

  • \[\begin{aligned} -2+\dfrac{v_1}{2}+\dfrac{v_2}{4}+7&=0 \end{aligned}\]
    - Combining super-nodes

Problem-5

  • Find the power of the dependent current source?

image

Solution-5

  • \[\begin{aligned} \dfrac{V_A-20}{5}+\dfrac{V_A}{5} & = 4 \\ \Rightarrow V_A & = 20~\mathrm{V} \end{aligned}\]
    Apply KCL at node-A
  • \[=20\times 4 = 80~\mathrm{W}\]
    Power delivered by the dependent source
image

Problem-6

  • Find the current through the 5\(\Omega\) resistor?

image

Solution-6

  • \[\begin{aligned} -2+\dfrac{V_1}{2}+\dfrac{V_1-V_2}{5}&=0 \end{aligned}\]
    Apply KCL at node-1
  • \[\begin{aligned} -\dfrac{V_1-V_2}{5}+4+\dfrac{V_2}{4}&=0 \end{aligned}\]
    Apply KCL at node-2
  • \[V_1 = 0.4~\mathrm{V} \quad V_2 = -8.7~\mathrm{V}\]
    On solving, we get

Problem-7

  • Find \(V1\) ?

image

Solution-7

  • \[\begin{aligned} 0.2V_1 +(10\times 10^{-3})+\dfrac{V}{10} - (2\times 10^{-3})+\dfrac{V_1}{1}&=0 \\ \Rightarrow V_1 &= 9.23~\mathrm{mV} \end{aligned}\]
    Applying KCL at node

Problem-8

  • Find the current through the 5m\(\Omega\) resistor?

image

Solution-8

  • \[\begin{aligned} &\dfrac{V}{4m\Omega}-(4~\mathrm{mA})+\dfrac{V}{3~m\Omega}-(10~mA)+\dfrac{V}{5~m\Omega}-5~mA =0\\ &\Rightarrow V = 11.54\times 10^{-6}~\mathrm{V} \end{aligned}\]
    Applying KCL at node
  • \[I = \dfrac{11.54\times 10^{-6}}{5\times 10^{-3}} = 2.31\times 10^{-3}\]
    resistor is Current through 5m

Problem-9

  • Find the power supplied by the dependent source?

image

Solution-9

  • \[i_1 = \dfrac{v_1}{2}\]
    Also
  • \[v_1 = -40~\mathrm{V} \quad v_2 = -75~\mathrm{V} \quad i_1 = -20~\mathrm{A}\]
    On solving
  • \[P = (3i_1)(v_2) = 4.5~\mathrm{kW}\]
    Power supplied by dependent source
  • \[\begin{aligned} -15+\dfrac{v_1-v_2}{1}+\dfrac{v_1}{2} & = 0 \end{aligned}\]
    Applying KCL at node
  • \[-3i_1+\dfrac{v_2}{3}+\dfrac{v_2-v_1}{1} = 0\]
    Applying KCL at node
image

Problem-10

  • Determine the value of \(V1\) ?

image

Solution-10

  • \[-(-8)-(-3)+\dfrac{v_1-v_2}{3}+\dfrac{v_1-v_3}{4}=0\]
    KCL at node
  • \[-3+\dfrac{v_2-v_1}{3}+\dfrac{v_2}{1}+\dfrac{v_3}{5}-25+\dfrac{v_3-v_1}{4}=0\]
    -Super-node analysis at
  • \[v_3-v_2 = 22\]
    Constraint equation at super-node
  • \[v_1 = 1.071~\mathrm{V} \quad v_2 =10.5 ~\mathrm{V} \quad v_3 = 32.5~\mathrm{V}\]
    On solving
Electric Circuits & Networks