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electric circuit(CE pp)
發問:
http://img177.imageshack.us/img177/9629/electric3.jpg 36.B, when resistance decrease, current increase. Why resistance remains unchanged? And why voltage increase? 37.D 40.A, why (2)and(3) incorrect 46.A, especially st(3), explain this statement Thank you for your help!
最佳解答:
(36) The rated current Io = V/r where V is the rated voltage and r is the resistance of the appliance The actual current flowing is Io/2 = V/(R + r) V/2r = V/(R + r) => R = r When V is doubled to 2V, the current will become 2V/(R+r) = 2V/2r = V/r = Io so B is correct. (37) The Ammeter indicates the total current through R and also the voltmeter (although the current through the voltmeter is small) (40) When S is closed,, the R3 and A3 reduces to a short circuit. The equivalent resistance of the total circuit is reduced. Therefore the current through R1 and R2 will increase. The current through a resistance is the ratio of the voltage to resistance. The voltage across R1 (and A1) is the same as the voltage across R2 (and A2). I1 = V/R1, I2 = V/R2 therefore I1/I2 = R2/R1 will not change no matter S is closed or not. When S is closed, all current will be directed to the short circuit path, and no current will flow through A3. (46) Equivalent resistance for X is given by 110^2/Rx = 40 or Rx = 110^2/40 = 302.5 Ohm Equivalent resistance for Y is given by 110^2/Ry = 80 or Ry = 110^2/80 = 151.25 Ohm When connected to 220V, the power for X will be 220^2/302.5 = 160W When X and Y are in series, the equivalent resistance is 302.5+151.25 = 453.75 Ohm Current = 220/453.75 = 0.4848A Power consumption for X = 0.4848^2 * 302.5 = 71.1W Power consumption for Y = 0.4848^2 * 151.25 = 35.6W
其他解答:FAD2A23AB937987B
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