Problem:

Could someone please give me hints or ideas of what method of circuit analysis I should use?
I'm having trouble starting this problem

 

Series parallel circuit with ohms law

 

Define \(I_0\) in terms of \(R_1\), \(R_2\), and \(I\) in both situations: switch open; switch closed. You can call them \(I_{0,open}\) and \(I_{0,closed}\).
\(
I_{0,closed} = I/2 = f(R, R_1) \\
I_{0,open} = I/3 = f(R, R_1, R_2) \\
\frac{I_{0,closed}}{I_{0,open}} = \frac{I/2}{I/3} = \frac{3}{2} = \frac{f(R, R_1)}{f(R, R_1, R_2)}
\)
This is basically the solution. Only one of the available answers will follow this relation.

Another way to look at it is to first visualize the circuit when the switch is closed: \(R_2\) disappears, and there are now only 2 parallel resistors. For 1/2 of the current to go into \(R_1\), or \(\frac{I_{0,closed}}{I} = \frac{1}{2}\), they must be equal. Then visualize the circuit when the switch is open: \(R_2\) appears, and the current decreases to \(I/3\). This means that current in the other branch increased to \(I \times 2/3\). Current is inversely proportional to resistance, or \(I \propto 1/R\), which means that the resistance in the switching branch is \(\frac{\left(\frac{1}{1/3}\right)}{\left(\frac{1}{2/3}\right)} = 2\) times larger than the resistance in the other branch (\(R\)). In other words, \(\frac{R_1 + R_2}{R} = 2\).