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.
Calculating total resistance for resistor network circuit
circuit analysisresistanceresistors
Best Answer
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
$$i_2(t) = \frac{R_2}{R_3 + R_4 + R_2}i_S(t) = \frac{R}{3R}i_S(t) = \frac{1}{3}i_S(t)$$