Todai Entrance Exam: Subject 2011 – Problem 1

(1) There is a short-cut at the ammeter position (reference), so:

    \[ A = \frac{E}{4R} \]

(2) 2R

(3) R_{TH} = R, V_{TH} = \frac{E}{2} (reference).

(4) A = \frac{E}{16R} (reference).

(5) In Figure 1:

    \[ A_{1} = \frac{E}{R}\left ( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \right ) \]

In Figure 2:

    \[ A_{2} = \frac{E}{R}\left ( \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} \right ) \]

(6) An error on a resistor of the first circuit will affect the value of only one bit, the current value is guaranteed to be correct as intended if that bit is off. An error on a resistor of the second circuit will affect the value of all bits, which will result in an erronous current value regardless of the input bits.


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