# Resistors-2

The natural successor to resistors-1. Some interesting strategies and shortcuts start to appear in this problem. We also need to start thinking in terms of Kirchhoffs laws, if not necessary applying them.

The first law says that at any given node $$\sum I_k = 0$$

The second law says that around any closed loop, $$\sum V_k =0$$

Solution

The answer is $R_{effective} = \frac{3}{2}\Omega$

To establish this, we need to consider two aspects of the system.

1. The square in the top left of the diagram
2. The potential difference between A & B
As the across each resistor is $1\times$ the current, the pd is numerically equal to the current across each component. The pd between A & B is therefore $(I_1 + I_2 + I_2 + I_1) = 2(I_1 + I_2)$, this is most easily established by travelling around the outside of the diagram. Looking at the loop in the top left corner, we know the pd around the loop is zero, so we get $$I_2 + I_2 - (I_1 - I_2) - (I_1-I_2) = 0 \Rightarrow 2I_1 = 4I_2 \Rightarrow I_2 = \frac{1}{2}I_1$$ Substitution of this into the earlier equation for the pd gives $V = 3I_1$ as the total current is $2I_1$, then $$R = \frac{V}{I} = \frac{3I_1}{2I_1} = \frac{3}{2}\Omega$$