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I'm looking to find the value of \$i_1\$, \$v\$, and the total power generated/absorbed.

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I started by applying KVL and KCL laws:

\$B: i_1=i_2+i_3\$

\$M_1: 1V=6i_2+5V+54k\Omega\$,

\$M_2: 8V=1.8k\Omega i_3-30i_1+6I_2\$

I tried to solve for \$i_1\$ usuing a matrix, but I didn't get anything close to right answer.

$$ \left[ \begin{array}{ccc|c} -30 &6 &1.8k &8 \\ 54k &6 &0 &-4 \\ 1 &-1 &-1 &0 \\ \end{array} \right] $$


There is no need for \$i_2\$ since the CCCS in the second branch is causing an integral multiple of \$i_1\$ to flow there and hence the middle \$6k\Omega\$ resistor has \$30+1=31i_1\$ flowing through it.

\$\text{KVL on }M_1:\$ \$ \begin{align} -5V+(54k\Omega)i_1-1V+(6k\Omega)(31i_1)=0\\ \therefore \quad i_1(54k\Omega+186k\Omega)=6\\ \therefore i_1=\frac{6}{240k\Omega}=25 \mu A \Longleftarrow \end{align}\$

Voltage across the central \$6k\Omega\$ resistor equals \$6k\Omega \times 31 \times 25\mu A=4.65V\$


\$\text{KVL on }M_2:\$ \$ \begin{align} 4.65V-8-(1.8k\Omega\times 30)i_1-\nu=0\\ \therefore \nu=-4.7 V \Longleftarrow \end{align}\$

Calculations for power dissipation are then easy to take up from this point onwards.


Your M1 is: 1V=6i2+5V+54kΩ

The first mistake is that 1V and 5V have the same polarity so your equation should be written as : 1V+5V=6i2+54kΩ

The second mistake is that you can Not add voltage values to resistance values, I think you forgot to multiply the 54K Resistor by its current to get the voltage across it and then you can insert the voltage value in the equation.

From KCL : The current in the 6K resistance is 31*i1

M1: "Left Loop"

-5 + 54*i1 + 6*(i1 + 30*i1) = 0

i1 = 48 mA

M2: The total loop or the outline loop

-5 + 54*i1 - 1 + v - 1.8*30*i1 + 8 = 0

V = -2 volts

I feel like there is a calculation mistake in my answer so, I'm not quite sure of my answer if it is wrong please tell me and I will try to solve it again


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