一條物理電路問題

有一個電阻值為50Ω的負荷連接4.5V的電池組, 串聯接駁一個或多個總共400歐姆既電阻器把負荷兩端的電壓調節為0.5V

.因爲
.負載電流 = I = V/R = 0.5 / 50 = 0.01
.調節電壓 = 4.5V - 0.5V = 4V
.降壓電阻 = R = V/I = 4 / 0.01 = 400

希望幫到你

負荷兩端的電壓調節為0.5V 時的電流:
V=IR
0.5=I(50)
I=0.01A
整個電路電阻值:
V=IR
4.5=0.01R
R=450Ω
所需電阻=450-50
=400Ω

let resistance of the load = RL = 50 ohm, and Voltage across the Load is 0.5V

According to Ohm's Voltage Law, V= IR, so 0.5 = I x 50
current I = 0.01 A or 10mA

Also, by ohm's Current law, current flowing in is equal to current flowing out. Therefore the circuit becomes I (R + RL) = 4.5V

Where I = 0.01A, RL = 50 ohm and the question is about the voltage drop across the R

Hence R = (4.5/0.01) - 50 = 450 - 50
= 400 ohm//

by ratio
R/50=4.0/0.5
R=400ohm
A resistor of 400 ohm should be connected in series, or resistors with equivalent of 400ohm should be added.