Question

how do you solve nodal analysis in which youre asked to find power dissipated in each resistor.
the resistors are labelled with S(which is power from my understanding) but that would change the formula for the nodal analysis which is v/r right? so how do i solve it?

Answer 1

Answer on Question #61197-Engineering-Electrical Engineering

How do you solve nodal analysis in which you're asked to find power dissipated in each resistor.

The resistors are labeled with S (which is power from my understanding) but that would change the formula for the nodal analysis which is v/r right? So how do i solve it?

Answer

The resistors are labeled with their conductance:

$$G = \frac {1}{R}.$$

I give you an example of such problem.

3.15 Apply nodal analysis to find i0 and the power dissipated in each resistor in the circuit.

Solution

Nodes 1 and 2 form a supernode so that $v_{1} = v_{2} + 10$ (1)

At the supernode, $2 + 6v_{1} + 5v_{2} = 3(v_{3} - v_{2}) \rightarrow 2 + 6v_{1} + 8v_{2} = 3v_{3}$ (2)

At node 3, $2 + 4 = 3(v_{3} - v_{2}) \rightarrow v_{3} = v_{2} + 2$ (3)

Substituting (1) and (3) into (2),

$$\begin{array}{l} 2 + 6 v _ {2} + 6 0 + 8 v _ {2} = 3 v _ {2} + 6 \rightarrow v _ {2} = - \frac {5 6}{1 1} = - 5. 1 \, \mathrm{V} \\ v _ {1} = v _ {2} + 1 0 = \frac {5 4}{1 1} = 4. 9 \, \mathrm{V} \end{array}$$ $$i _ {0} = 6 v _ {1} = 2 9. 4 5 A$$ $$P _ {6 5} = v _ {1} ^ {2} G = \left(\frac {5 4}{1 1}\right) ^ {2} 6 = 1 4 4. 6 W$$ $$P _ {5 5} = v _ {2} ^ {2} G = \left(- \frac {5 6}{1 1}\right) ^ {2} 5 = 1 2 9. 6 W$$ $$P _ {3 5} = (v _ {2} - v _ {3}) ^ {2} G = (2) ^ {2} 3 = 1 2 W$$

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