By inspection, write the mesh-current equations for the circuit in Fig. 3.29.

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Accepted Answer

We have five meshes, so the resistance matrix is 5 by 5. The diagonal terms, in ohms, are:

[latex]\begin{gathered}R_{11}=5+2+2=9, \quad R_{22}=2+4+1+1+2=10 \R_{33}=2+3+4=9, \quad R_{44}=1+3+4=8, \quad R_{55}=1+3=4\end{gathered}[/latex]

The off-diagonal terms are:

[latex]\begin{gathered}R_{12}=-2, \quad R_{13}=-2, \quad R_{14}=0=R_{15} \R_{21}=-2, \quad R_{23}=-4, \quad R_{24}=-1, \quad R_{25}=-1 \R_{31}=-2, \quad R_{32}=-4, \quad R_{34}=0=R_{35} \R_{41}=0, \quad R_{42}=-1, \quad R_{43}=0, \quad R_{45}=-3 \R_{51}=0, \quad R_{52}=-1, \quad R_{53}=0, \quad R_{54}=-3\end{gathered}[/latex]

The input voltage vector v has the following terms in volts:

[latex]\begin{gathered}v_{1}=4, \quad v_{2}=10-4=6 \v_{3}=-12+6=-6, \quad v_{4}=0, \quad v_{5}=-6\end{gathered}[/latex]

Thus the mesh-current equations are:

[latex]\left[\begin{array}{rrrrr}9 & -2 & -2 & 0 & 0 \-2 & 10 & -4 & -1 & -1 \-2 & -4 & 9 & 0 & 0 \0 & -1 & 0 & 8 & -3 \0 & -1 & 0 & -3 & 4\end{array} ight]\left[\begin{array}{l}i_{1} \i_{2} \i_{3} \i_{4} \i_{5}\end{array} ight]=\left[\begin{array}{r}4 \6 \-6 \0 \-6\end{array} ight][/latex]

From this, we can obtain mesh currents [latex] i_{1}, i_{2}, i_{3}, i_{4}, ext { and } i_{5}[/latex].

Question 3.9: By inspection, write the mesh-current equations for the circ...

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