Electronics - Designing a ciruit

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Homework Help Overview

The discussion revolves around designing a circuit with an ideal power supply of 15 V and a load of 5 kilo ohms, aiming for a current of approximately 2 mA. The original poster expresses confusion regarding the role of the load in the circuit design.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster considers using two resistors of 3.75 kilo ohms each but questions the significance of the 5 kilo ohm load. Another participant explains the nature of a load and its role in consuming electrical energy. A suggestion is made to use two resistors in series with the load to achieve the desired current.

Discussion Status

The discussion is active, with participants exploring different interpretations of the load's role and proposing circuit configurations. While some guidance has been provided regarding the use of resistors in series, there is no explicit consensus on the final circuit design.

Contextual Notes

Participants are navigating the implications of using an ideal power supply and the characteristics of the load in their circuit design. The original poster's uncertainty about the load's function indicates a need for further clarification on these concepts.

kel
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Homework Statement



Given an ideal power supply of 15 V and a load of 5 kilo ohms, design a circuit using 2 resistors in which the current flowing through the load is close to 2mA

Homework Equations



None given (I'm assuming V=IR)


The Attempt at a Solution




I'm a bit confused with this one, I understand that maybe I could have 2 resistors of 3.75 kilo ohms each to cause the current to me 2mA, but I'm unsure of what role the '5 kilo ohm load' has.

Can anyone explain to me what this load is and what role it plays? I can't see it being internal resistance due to the power supply being 'ideal'.

Thanks
Kel
 
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A load could be a bunch of things, say a speaker or a lightbulb or an appliance. They consume electrical energy and can dissipate it as sound, light or heat...whatever the case may be. I believe you need to consider the resistance of the load in your design.
 
So basically I could just have 2 resistors in series with a total resistance of 2.5 kilo ohms and the 5 k'ohm load.

Thus, using I = v/R I'd get the 2mA that is expected!

Does that make sense??
 
Thread moved from Advanced Physics to Intro Physics.

kel, yes your answer is correct. By placing the 2.5 kOhms in series with the 5.0 kOhms, that will give you a 2mA current supplied by the 15V power supply.
 

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