Solving Electric Current Homework

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SUMMARY

The discussion focuses on solving a homework problem related to electric current involving a battery with electromotive force (emf) E and internal resistance r. The user attempts to apply Kirchhoff's loop rule to derive equations for the emf and internal resistance, ultimately determining the resistance R needed to achieve a current of 3 amperes. The correct approach involves using the equation E = Ir + I R, clarifying the relationship between current, voltage, and resistance.

PREREQUISITES
  • Understanding of Kirchhoff's loop rule
  • Familiarity with Ohm's Law (V = IR)
  • Knowledge of electromotive force (emf) and internal resistance
  • Basic concepts of electric circuits
NEXT STEPS
  • Study the application of Kirchhoff's laws in circuit analysis
  • Learn about the characteristics of batteries and their internal resistance
  • Explore the relationship between voltage, current, and resistance in series and parallel circuits
  • Investigate the concept of maximum current in electrical circuits
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone looking to deepen their understanding of electric circuits and battery behavior.

Kandycat
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Homework Statement


2crulcp.jpg


A battery with emf, [tex]E[/tex], and internal resistance, r, is connected to a variable resistance, R, at points X and Y, as shown above on the left. Varying R changes, both the current I and the terminal voltage Vxy. The quantities I and Vxy are measured for several values of R and the data are plotted in a graph, as shown above the right.

(a) Determine the emf of the battery.

(b) Determine the internal resistance of the battery.

(c) Determine the value of the Resistance that will produce a current I of 3 amperes.

(d) Determine the maximum current that the batter can produce.

Homework Equations



V=IR

The Attempt at a Solution



I attempted to use Kirchhoff's loop rule. I'm not sure if I should be using it or not.

(a) [tex]E[/tex] - Ir - R = 0
[tex]E[/tex] = Ir + R

(b) r = ([tex]E[/tex] - R)/I

(c) V = IR
R = V/I = 3V/3A = 1 ohms

(d) I have no idea what to do.

Help?
 
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Kandycat said:

The Attempt at a Solution



I attempted to use Kirchhoff's loop rule. I'm not sure if I should be using it or not.

(a) [tex]E[/tex] - Ir - R = 0
[tex]E[/tex] = Ir + R

Instead of "R", that term should be "I R", the voltage across resistor R.
 

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