Resistor Values

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

  1. tquiva

    Thread Starter Member

    Oct 19, 2010


    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
    Series parallel circuit with ohms law
  3. tyblu


    Nov 29, 2010
    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}.
    <br />
I_{0,closed} = I/2 = f(R, R_1) \\<br />
I_{0,open} = I/3 = f(R, R_1, R_2) \\<br />
\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)}<br />
    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.