Ls 發問於 科學及數學物理學 · 5 年前

Circuit question!!?

Isn't there a short circuit in one of the wire PQ, so the lamp should remain very bright thorough the process? Why is the answer D?

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1 個解答

  • 天同
    Lv 7
    5 年前

    The circuit shown in the given diagram is actually two variable resistors connected in parallel, which is then connected in series to the battery and the light bulb. When the slider moves, one variable resistor decreases, the other will increase. The sum of the resistances of the two variable resistors must be equal to the resistance of the rheostat coil).

    There is no short circuit, except when slider X touches the ends of the rheostat coil at P and Q. When X is at some distance from P (as shown on the diagram), current flows from Q along the upper brass bar to X, where the flow of current divides into two paths, one through the coil to P (i.e. to the left) and back to the battery, the other through the coil to the lower pin at Q (i.e. to the right), and then back to the battery (through the pin at P). Therefore, the circuit is two variable resistors in parallel.

    Let the resistance of the rheostat coil be R. When the slider X is at the middle of the coil, the resistance to the left and right sides of the coil is R/2. The equivalent resistance of the two resistances in parallel is thus (R/2)/2 = R/4.

    When slider X is at either end of the rheostat coil, current can by-pass the rheostat coil. Under this situation, the equivalent resistance of the "rheostat circuit" is zero and current through the light bulb is maximum, the bulb glows with maximum brightness.

    Therefore, when slider X moves from P to Q, the equivalent resistance of the "rheostat circuit" varies from zero (occurs when slider X is at P) to a maximum of R/4 when X is at the middle of the rheostat coil and then decreases to zero when X reaches Q. As the equivalent resistance varies, the current thus decreases from maximum to minimum, and then back to maximum again (remember that current varies inversely as the equivalent resistance of the circuit). Because brightness of the bulb decreases with decrease of current through it and vice versa. The bulb brightness then decreases from its maximum brightness to a minimum and then back to maximum again.