# 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.

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

## 2 Answers

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.

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