# Total Power Absorbed with KVL and KCL

simulate this circuit – Schematic created using CircuitLab

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

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 $6k\Omega \times 31 \times 25\mu A=4.65V$

Hence

$\text{KVL on }M_2:$ \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.

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