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問題係呢到： http://www.mrleung.com/upload/i/13470303191491.png 個R會係幾多呢？
最佳解答:
It is obvious that R should be connected as shown in fig.2b. In fig.2a, R is connected in parallel to the battery, thus the voltage across R2 doesn't change. Its power dissipation is 40^2/20 w = 80 w, which exceeds the specified limit. In fig.2b, when R is connected in series, the voltage drop across R causes a reduction in voltage drop across R2, leading to a decrease of power on R2. If the power on R2 is limited to 40 w, then voltage across R2 = square-root[40 x 20] v = 28.28 v Voltage across R = (40 - 28.28) v = 11.72 v I2 = 28.28/20 A = 1.414 A I1 = 28.28/(10+30) A = 0.707 A Hence, current through R = (1.414 + 0.707) A = 2.121 A Therefore, R = 11.72/2.121 ohms = 5.53 ohms
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一題電學題目發問:
問題係呢到： http://www.mrleung.com/upload/i/13470303191491.png 個R會係幾多呢？
最佳解答:
It is obvious that R should be connected as shown in fig.2b. In fig.2a, R is connected in parallel to the battery, thus the voltage across R2 doesn't change. Its power dissipation is 40^2/20 w = 80 w, which exceeds the specified limit. In fig.2b, when R is connected in series, the voltage drop across R causes a reduction in voltage drop across R2, leading to a decrease of power on R2. If the power on R2 is limited to 40 w, then voltage across R2 = square-root[40 x 20] v = 28.28 v Voltage across R = (40 - 28.28) v = 11.72 v I2 = 28.28/20 A = 1.414 A I1 = 28.28/(10+30) A = 0.707 A Hence, current through R = (1.414 + 0.707) A = 2.121 A Therefore, R = 11.72/2.121 ohms = 5.53 ohms
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