Reducing voltage with two resistors

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Homework Statement



An electric potential of 5.0 V is required to run certain computer chips. A 6.0 V battery may be used to do this but it must be connect to two resistors in series. Supposing one has a resistance of 330 Ohms what should the other be? (The computer chip will be driven by the voltage across only one of the two resistors.)

Homework Equations


Req= R1+R2+...
Voltage=current x req

The Attempt at a Solution


I realize that equivalent resistance for series circuits just involves simple addition, but I don't know how to tie that in with voltage. Initially I tried solving for current (330 ohms(I)= 6V) and got .018 Amperes. plugged that into 5 volts to find the req (5v= R(.018)) and came out with 277. subtracted that from 330 and got 53 which is wrong. ?
 
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The question is awkwardly worded, but the key is in the parentheses at the end. Picture the two resistors in series. Picture the computer chip in parallel with the second resistor. Finally assume that the computer chip draws very little current compared to what flows through the two resistors. (A terrible assumption if this were the real world). Can you find a value for the second resistor such that the voltage drop across the second resistor is 5 volts?
 
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Moderator note: I've re-named your thread to: Reducing voltage with two resistors. Thread titles must be descriptive of the question or problem being asked, not a plea for help. Refer to the pinned thread: Guidelines for students and helpers pinned at the top of the thread list for more information.
 
Cutter Ketch said:
The question is awkwardly worded, but the key is in the parentheses at the end. Picture the two resistors in series. Picture the computer chip in parallel with the second resistor. Finally assume that the computer chip draws very little current compared to what flows through the two resistors. (A terrible assumption if this were the real world). Can you find a value for the second resistor such that the voltage drop across the second resistor is 5 volts?
thanks! the wording of the question did confuse me, so much appreciated