# Calculating total resistance for resistor network circuit

circuit analysisresistanceresistors

I'm very lost in my ECE class right now and I was hoping someone could help me out and explain what they're doing. I was asked to calculate the total resistance. After getting that, I was asked to calculate the \$i_2(t)\$ in terms of \$R\$ and \$i_S(t)\$. I am so lost and any help in explaining this to me would be greatly appreciated.

To find the equivalent resistance across terminals \$a\$ and \$b\$ first set the independent sources to zero. In this case you have a current source so when it is set to zero it is equivalent to an open circuit (if it was a voltage source then it would be equivalent to a short circuit). Your circuit now looks like this:

simulate this circuit – Schematic created using CircuitLab

\$R_1\$ is in series with an open circuit so it can be ignored. You have three remaining resistors. \$R_2\$ is in series with \$R_3\$ so their combined resistance is \$2R\$. This resistance is in parallel with \$R_4\$ so

$$R_{eq} = 2R \parallel R_4 = 2R \parallel R = \frac{2R^2}{2R + R} = \frac{2}{3}R$$

To find \$i_2(t)\$ you can use a current divider. Since \$a\$ and \$b\$ are open terminals the current through \$R_3\$ is simply \$i_2(t)\$, and the equivalent resistance in this path is \$R_3 + R_4 = 2R\$. By KCL at the node between \$R_1\$, \$R_2\$, and \$R_3\$:

$$i_S(t) = i_1(t) + i_2(t)$$

The current divider at this node gives

$$i_2(t) = \frac{R_2}{R_3 + R_4 + R_2}i_S(t) = \frac{R}{3R}i_S(t) = \frac{1}{3}i_S(t)$$