Solving Variable Resistance Circuit: Ohm's Law

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Homework Help Overview

The problem involves a circuit with a variable resistance of 100 ohms and a voltage drop supplied by a battery. The original poster seeks to determine the resistance required for two different voltage drops: 12V and 6.2V, using Ohm's Law.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply Ohm's Law but expresses uncertainty about solving for two unknowns. Some participants question the completeness of the information provided, particularly regarding the circuit's configuration and the presence of a diagram.

Discussion Status

Participants are exploring different interpretations of the problem, with one suggesting a calculation based on assumed current. There is a lack of consensus on the correct approach, and some participants express doubt about the reliability of their calculations.

Contextual Notes

There is mention of a diagram that may be relevant to understanding the circuit's configuration, but it is not provided. The original poster notes that they have only learned Ohm's Law, which may limit their approach to the problem.

Warrzie
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Homework Statement


"A circuit has a variable resistance (total 100 ohms) across which is a voltage drop of 12 V supplied by a battery. What must be its resistance if the voltage is to be A) 12V and B) 6.2V?"

Homework Equations


V=IR


The Attempt at a Solution


We haven't learned anything besides Ohm's Law for these equations, so I tried plugging in different values to try and dial in the answers, but it didn't work. Is there something I'm missing or am I being asked to solve for two unknowns? Hints appreciated.
 
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Is that all you have? Diagram? I am assuming total resistance is 100 ohms so for 12 v you have 8.5 amps. Then for 12V, you need a resistance of 12/8.5 but I don't trust this answer. If its a serial pair of resistors where Rx and Ry are 100 ohms and you are being asked the voltage drop across Rx, i think the above is good.
 
Last edited:
drawing a blank on the diagram, literally
 

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