Sample Problem: Ohm's Law and Power 1

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Do curso por Georgia Institute of Technology

Linear Circuits 1: DC Analysis

103 classificações

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

103 classificações

This course explains how to analyze circuits that have direct current (DC) current or voltage sources. A DC source is one that is constant. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

Na lição

Module 1: Introduction

This module reviews background material from physics on the basic properties of electricity and electrical circuits.

Sample Problem: Ohm's Law and Power 13:39

Sample Problem: Ohm's Law and Power 22:52

Sample Problem: Power 15:34

Sample Problem: Power 23:07

Sample Problem: Power 31:40

Sample Problem: Power 41:57

Sample Problem: Voltage and Energy1:10

Sample Problem: Electric Current and Charge 11:04

Sample Problem: Electric Current and Charge 20:53

Sample Problem: Voltage Division1:24

Conheça os instrutores

Dr. Bonnie H. Ferri
Professor
Electrical and Computer Engineering
Dr. Joyelle Harris
Director of the Engineering for Social Innovation Center
School of Electrical and Computer Engineering
Dr Mary Ann Weitnauer

The topic of this problem is Ohm's Law and

also using the well-known equation for Power, as well.

The problem is determine the resistance R and the voltage source V sub s.

So we have a current source,

on the left-hand side of the circuit that also has a voltage associated with it.

And that's the V sub S that we're looking for.

To solve this problem, you have to use the passive sign convention in order

to be able to assign the polarities across a voltage drop of the elements.

So in this case, we have a voltage drop across the resistor that is

unknown as far as the polarity goes.

So we first assign a direction for the current.

And then, knowing the direction for the current that we've assigned,

we can then assign the polarity across the resistor.

In the passive sign convention, the current is assumed to enter the positive

side of the voltage polarity across the element.

So if we assign the current to go clockwise around this single loop circuit,

then the voltage drop across the resistor,

P sub R is going to be, as shown in the circuit now, and

ultimately we're going to decide what the resistance is and

what the voltage source of this V sub S is?

So we know that using the equation for

power, that power = I squared R.

And we're given the power in this problem.

The power in the resistor is equal to 80 milliwatts.

We're also given the current, the current is equal to 4 milliamps.

And the problem is on the left-hand side as a current source.

So we have everything we need in order to solve for the resistance.

So the resistance is going to be equal to P/I squared,

and P=80mW, it's positive,

I=4mA I'm going to square that and ultimately we end up with a resistance,

That's equal to 5 kiloohms.

So that solves for our resistance.

The next thing we need to find,

is we need to find the voltage associated with the current source.

So it's the source voltage across the current source.

We can do this because we know that all that voltage

that's across the voltage source is also the voltage across the resistor.

It's the only other element in the circuit.

So if we could solve for the voltage across the resistor, then we know that

that is going to be equal to the voltage across the current source.

So the voltage across the resistor,

which is equal to V sub S, is going to be equal

to the current times the resistance,

and so the current is 4 mA.

We calculated the resistance from a previous part of the problem, and so

we end up with a voltage which is equal to 20 volts.

So the voltage associated with the source is going to be 20 volts.

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