Compared to an ideal battery, by what percentage does the battery's internal resistance reduce the potential difference across the resistor in the figure (Part A 1 figure) 20 Ω?
Express your answer using two significant figures.
http://session.masteringphysics.com/problemAsset/1075268/3/32.Ex18.jpg
Thank you for your time and consideration!!!!
Help!!! Ideal Battery? Simple Circuit??? Help!?
2008-07-29 3:08 pm
回答 (3)
2008-07-29 3:20 pm
✔ 最佳答案
I = E/R = 15/31 = 0.484 ampsDrop across 1 ohm resistor is E = IR = 0.484 volts.
Percentage drop in battery voltage is
0.484/15 = 0.0323 = 3.2%
This percentage drop applies throughout the circuit, so it is the drop across the 20 ohm resistor.
.
2008-07-29 6:04 pm
regardless you get 31 ohms resistance plus what ever resistance the battery inherently has.........tom
2008-07-29 3:44 pm
when current flows through the circuit I=V/R
that is 15/31=0.4838A
since the circuit is in series current flows all the resistance are in same so current for 20 ohms resistor is also 0.4838
the drop across 20 ohms resistor is V=IR
voltage drop in 20ohm resistor=0.4838*20=9.68V
that is 15/31=0.4838A
since the circuit is in series current flows all the resistance are in same so current for 20 ohms resistor is also 0.4838
the drop across 20 ohms resistor is V=IR
voltage drop in 20ohm resistor=0.4838*20=9.68V
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