# Why do i get three different methods to calculate the same current ,but get three different current values?

I want to find the current $$\I_R\$$,and i used three methods,and i got three different $$\I_R\$$ values,i hope someone can tell me where am i wrong!

simulate this circuit – Schematic created using CircuitLab

method 1 & 2 : There are two circuits as below,the main difference between them is the direction of $$\I_2\$$,which are opposite!

Left hand circuit :

$$\I_R=I_1=3\frac{(2+3)}{(2+3)+(1+R_L)}=3\frac{5}{6+R_L}=\frac{15}{6+R_L}\$$

Right hand circuit :

$$\I_R=I_1=3+I_2=3+1\frac{3}{3+(2+1+R_L)}=3+\frac{3}{6+R_L}=\frac{21+3R_L}{6+R_L}\$$

simulate this circuit

method 3:superposition

$$\I_R=I_{R1}+I_{R2}+I_{R3}=0+[3\frac{(2+3)}{(2+3)+(1+R_L)}]+[1\frac{3}{3+(2+1+R_L)}]=\frac{18}{6+R_L}\$$

simulate this circuit

If we use the maximum power Maximum Power Transfer Theorem,we can know $$\R_L=6Ω\$$.However,now i have a big problem,that is,we can see that when i replace $$\R_L\$$ with $$\6Ω\$$ in the current from method 1 to method 2,we can find that we have three different $$\I_R\$$ values!!

I believe that $$\I_R\$$ value by using method 3 is correct,but i don't know where am i wrong in my method 1 and method 2,can anyone tell me the mistake i made?

• You should add units to your equations. The assumption $I_R=I_1=3A \frac{(2 \Omega+3\Omega)}{(2\Omega+3\Omega)+(1\Omega+R_L)}$ is incorrect. The current of 3A isn't divided across R3,R2, R4 and RL like that, because it does not take into account the contribution of the 1A current. Commented Apr 29, 2020 at 7:34
• You should add units to your equations that will help as a "sanity check" so that you'll notice when you're making a mistake and trying to add a voltage to a current for example. Commented Apr 29, 2020 at 7:39

Method 3 is correct.

$$\I_R=I_{R1}+I_{R2}+I_{R3}=\color{orange}{0\text{ A }}+\color{magenta}{[3\text{ A }\frac{(2\Omega+3\Omega)}{(2\Omega+3\Omega)+(1\Omega+R_L)}]} +\color{blue}{[1\text{ A }\frac{3\Omega}{3\Omega+(2\Omega+1\Omega+R_L)}] } =\frac{18\Omega}{6\Omega+R_L}\text{ A}\$$

That method also shows what goes wrong in method 1 and 2:

In the Left hand circuit:

$$\I_R=I_1=3 \text{ A } \frac{(2\Omega+3\Omega)}{(2\Omega+3\Omega)+(1\Omega+R_L)}\$$

The current of 3A isn't divided across $$\R_3\$$, $$\R_2\$$, $$\R_4\$$ and $$\R_L\$$ like that, because it does not take into account the contribution of the 1A current.

You can see it comparing it to the solution of method 3: Only the contribution of the 3A current source (magenta part) is taken into account by this method 1, not the contribution of the 1A current source (blue part).

Right hand circuit:

$$\I_R=I_1=3\text{ A }+I_2=3\text{ A }+1\text{ A }\frac{3\Omega}{3\Omega+(2\Omega+1\Omega+R_L)}\$$

The problem here is that the 3A current is not only running through $$\R_9\$$ and $$\R_L\$$, but also through $$\R_8\$$ and $$\R_7\$$.

Comparing to method 3 you can see that only $$\ \frac{(2\Omega+3\Omega)}{(2\Omega+3\Omega)+(1\Omega+R_L)}\$$ part of the 3A current flow through $$\R_L\$$.

Regarding method 1 correct approach would be:

Using the left hand circuit:

The voltage across $$\R_2\$$ is given by $$\V_{R_2}= (i_2+1\text{ A }) \cdot R_2 \$$
The voltage across $$\R_3\$$ is given by $$\V_{R_3}=i_2 \cdot R_3 \$$

The sum of these voltages is across $$\R_1\$$ and $$\R_L\$$, so the current through $$\R_L\$$ is given by $$i_{R_L} = \frac{ V_{R_2}+V_{R_3} }{R_1+R_L} = \frac{ (i_2+1\text{ A }) \cdot R_2 + i_2 \cdot R_3 }{R_1+R_L}$$

According KCL, $$\i_2+i_{R_L}=3\text{ A }\$$. Substituting for $$\i_2\$$ this yields

$$i_{R_L} = \frac{ (3\text{ A }-i_{R_L} +1\text{ A }) \cdot R_2 + (3\text{ A }-i_{R_L}) \cdot R_3 }{R_1+R_L}$$

$$i_{R_L} (R_1+R_L) = 4\text{ A } \cdot R_2 - i_{R_L} \cdot R_2 + 3\text{ A } \cdot R_3 - i_{R_L}\cdot R_3$$

$$i_{R_L} (R_1+R_2+R_3+R_L) = 4\text{ A } \cdot R_2 + 3\text{ A } \cdot R_3$$

$$i_{R_L} = \frac{ 4\text{ A } \cdot R_2 + 3\text{ A } \cdot R_3 }{R_1+R_2+R_3+R_L}$$