Current and voltage dividers with multiple sources

Question

In a practice problem for my ECE class I am told to use a voltage divider to find \$v_1\$ in the circuit below

And to use a current divider to find the \$i_1\$ in this circuit

I am given the answer to each problem as these are just for practice, and so I know that \$V_1=2V, i_1 = 27mA\$ And I know the formula for a voltage divider or current divider as

For n resistors in series: \$V_n = V_{total}(\frac{R_n}{R_1+R_2 + ... + R_n})\$

For n resistors in \$\parallel\$ : \$i_n = i_{total}(\frac{G_n}{G_1 + G_2 + ... G_n})\$ Where \$G\$ is conductance is \$\frac1R\$

I have tried to use these formulas, but I don't know what to do when there are multiple sources in a circuit. I could use a KCL or KVL to analyze the circuit but I'm asked specifically to use a divider.

Answer 1

For the first circuit diagram, did you notice that the 3 Ω resistor is short circuited, and hence the + terminal of the ideal 18 V source is connected to the top of the 2 Ω resistor? Knowing this, you can use the voltage divider on the 2 Ω and 10 Ω resistors : $$v_1=\frac{2}{10+2}\times (18-6)\:\text{V}=2\:\text{V}$$

For the second circuit diagram, the 36 mA source only feeds the parallel 1.2 kΩ and 3.6 kΩ resistors. The remainder of the components are just hanging off the lower terminal of the 36 mA source, and have no return path to the top of this current source (easy to see if you re-draw the circuit and move the 3.6 kΩ resistor up, to join the node just below the 36 mA source). Hence the current divider can be used: $$i_1=\frac{3.6}{1.2+3.6}\times 36\:\text{mA} =27\:\text{mA}$$