Pi Circuit Analysis

There is no need to work with formulas for equivalent resistance.

Full circuit: 10 $A$ at 6 $V$ .

The leftmost resistor (between $j_1$ and $j_3$ ) is in parallel with the rest. It has 6 $V$ and therefore 6 $A$ of current. For the rest of the circuit, this leaves 4 $A$ at 6 $V$ .

The top resistor (between $j_1$ and $j_2$ ) is in series with the remaining two. It has 4 $A$ of current and therefore a 4 $V$ potential drop. For the rest of the circuit, this leaves 4 $A$ at 2 $V$ .

The right 1 $\Omega$ resistor is parallel with the load. It has 2 $V$ and therefore 2 $A$ of current. This leaves 2 $A$ and 2 $V$ for the load.

The resistance of the load is therefore $\dfrac{2 V}{ 2 A} = \boxed{1 \Omega} .$

The resistors between $j2,j4$ and $c2,c4$ are in parallel. Their resultant $\dfrac{R_L}{1+R_L}$ is in series with the resistor between $j1,j2$ . This gives $R=1+\dfrac{R_L}{1+R_L}$ .
Now, the voltage across the resistor between $j1,j3$ is $6V$ . Therefore, the current through it will be $\dfrac{6V}{1 \Omega} = 6A$ . This implies the current thought the rest of the circuit of effective resistance $R$ will be $A$ .
Hence, we can write,
$\begin{aligned} V &=& iR \\ 6 &=& 4 \left(1 + \dfrac{R_L}{1+R_L} \right) \\ \frac{R_L}{1 + R_L} &=& \frac{1}{2} \\ R_L &=& \boxed {1 \Omega}. \end{aligned}$

Equivalent resistance is 0.6 ohm. To get 0.6 with leftmost 1 ohm in parallel, other resistance should be 1.5, which is after combining with middle 1 ohm, so we get 0.5 ohm combining with rightmost 1 ohm in parallel. This means R is 1 ohm.

The equivalent resistance between $c_1$ and $c_3$ is given by:

$\begin{aligned} R_{eq} & = 1 || \left(1+1||R_L \right) \\ & = 1 || \left(1+\frac {R_L}{1+R_L} \right) \\ & = 1 || \left(\frac {1+2R_L}{1+R_L} \right) \\ & = \frac {\frac {1+2R_L}{1+R_L}}{1+\frac {1+2R_L}{1+R_L}} \\ & = \frac {1+2R_L}{2+3R_L} \end{aligned}$

Now, we have:

$\begin{aligned} R_{eq} & = \frac 6{10} \\ \implies \frac {1+2R_L}{2+3R_L} & = 0.6 \\ 1 + 2R_L & = 1.2 + 1.8 R_L \\ 0.2R_L & = 0.2 \\ \implies R_L & = \boxed{1} \ \Omega \end{aligned}$