Resistor Values

Discussion in 'Homework Help' started by tquiva, Dec 16, 2010.

Oct 19, 2010
176
1
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

2. R!f@@ AAC Fanatic!

Apr 2, 2009
9,616
1,082
Series parallel circuit with ohms law

3. tyblu Member

Nov 29, 2010
199
16
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, $\frac{R_1 + R_2}{R} = 2$.