標題:
hkcee mc(current and voltage)
發問:
最佳解答:
1. The figures in the diagram are not clear. I presume the battery voltage is 6v Equivalent resistance of pararllel circuit = 4 x (3+1)/(4+(3+1)) ohms = 2 ohms Total resistance = (4+2) ohms = 6 ohms Main current = 6/6 A = 1 A Current through 3-ohm resistor = 1/2 A = 0.5 A Hence, potential difference across XY = 0.5 x 3 v = 1.5 v 3. Use: power = voltage^2/resistance New power delivered by the heater P = Po(100/200)^2 where Po is the power when the heater is connected to 200v source. hence, P = Po/4 Energy needed to boil the wayer = (Po).T Hence, new time required = (Po).T/P = (Po).T/(Po/4) = 4T 7. After S is closed, total resistance of circuit = (10+20) ohms = 30 ohms current = 3/30 A = 0.1 A Hence, voltage across 20-ohm resistor = 0.1 x 20 v = 2 v Therefore, reading of V1 = 0 v (because voltmeter V1 is not connected across any resistor) , reading of V2 = 2v
其他解答:
1. B V = (6-4)* 3/(3+1) = 2 * 0.75 = 1.5 V 3. E P = E / t V^2 / R = E / t E = 40000 / R * T As energy required to boiled the water and the resistance of the kettle remain unchanged 40000 / R * T = 100^2 / R * t (t is the time required to boil the water for 100V source) 40000 T = 10000t t = 4T 7. A When switch S is closed, no current flow through V1 as the resistance of the path V1 is higher then that of the path with switch S. Current in the circuit = V / R = 3 / (10+20) = 0.1 A Reading of V2 = Voltage across 20 ohm resistor = IR = 20 * 0.1 = 2V|||||#32 V=IR 6=I [4+(1/4 + 1/4)^-1] I=2 A V=IR V=(1)(2) =2 V=2 x(3/4) =1.5V ans:B #35 E=Pt =V^2 T/R =200^2T/R E=V'^2 T'/R =100^2 T'/R T'=4T ans:E #33 V1=0 V=IR 3=I(10+20) I=0.1 A V=IR =0.1(20) =2 V ans:A
hkcee mc(current and voltage)
發問:
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圖片參考：http://imgcld.yimg.com/8/n/HA08059003/o/701203220079713873382320.jpg 求以上MC的步驟。謝謝最佳解答:
1. The figures in the diagram are not clear. I presume the battery voltage is 6v Equivalent resistance of pararllel circuit = 4 x (3+1)/(4+(3+1)) ohms = 2 ohms Total resistance = (4+2) ohms = 6 ohms Main current = 6/6 A = 1 A Current through 3-ohm resistor = 1/2 A = 0.5 A Hence, potential difference across XY = 0.5 x 3 v = 1.5 v 3. Use: power = voltage^2/resistance New power delivered by the heater P = Po(100/200)^2 where Po is the power when the heater is connected to 200v source. hence, P = Po/4 Energy needed to boil the wayer = (Po).T Hence, new time required = (Po).T/P = (Po).T/(Po/4) = 4T 7. After S is closed, total resistance of circuit = (10+20) ohms = 30 ohms current = 3/30 A = 0.1 A Hence, voltage across 20-ohm resistor = 0.1 x 20 v = 2 v Therefore, reading of V1 = 0 v (because voltmeter V1 is not connected across any resistor) , reading of V2 = 2v
其他解答:
1. B V = (6-4)* 3/(3+1) = 2 * 0.75 = 1.5 V 3. E P = E / t V^2 / R = E / t E = 40000 / R * T As energy required to boiled the water and the resistance of the kettle remain unchanged 40000 / R * T = 100^2 / R * t (t is the time required to boil the water for 100V source) 40000 T = 10000t t = 4T 7. A When switch S is closed, no current flow through V1 as the resistance of the path V1 is higher then that of the path with switch S. Current in the circuit = V / R = 3 / (10+20) = 0.1 A Reading of V2 = Voltage across 20 ohm resistor = IR = 20 * 0.1 = 2V|||||#32 V=IR 6=I [4+(1/4 + 1/4)^-1] I=2 A V=IR V=(1)(2) =2 V=2 x(3/4) =1.5V ans:B #35 E=Pt =V^2 T/R =200^2T/R E=V'^2 T'/R =100^2 T'/R T'=4T ans:E #33 V1=0 V=IR 3=I(10+20) I=0.1 A V=IR =0.1(20) =2 V ans:A
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