과목:전기회로(목 1,2 교시)

P82~P86 8문제중 7문제 14번)Six volts are applied across a resistor. A current of 2 mA is measured. What is the value of the resistor? solve) I=V/R, R=V/I 그러므로 R=6/(2) =3000 21번)Convert the following the microwatts: (a)2W (b)0.0005W (c)0.25mW (d)0.00667mW solve) (a)2=2000000W (b)0.0005=500W (c)0.25W=250W (d)667W=6.67W 28번)A 50 resistor is connected across the terminal of a 1.5V battery. What is the power dissipation in the resistor? solve) P=/R 이므로 =2.25/50 =0.09W 35번)For each circuit in Figure 3-29, assign the proper polarity for the voltage drop across the resistor.(죄송합니다. 그림은 실지 못했습니다.) solve) (a)위쪽에 위치한다. (b)아래쪽에 위치한다. (c)오른쪽에 위치한다. 42번)A certain power supply provides a continuous 2W to a load. It is operating at 60% efficiency. In a 24h period. how many kilowatt-hours does the power supply use? solve) 2W=0.002kW 효율이 60% 쓴시간은 24시간이다. 그러므로 0.002kW24h0.6 = 0.0288kW 49번) If you wish to increase the amount of current in a resistor from 100mA to 150mA by changing the 20V …
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