Resistor circuit, potential drop and time

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SUMMARY

The discussion focuses on calculating the potential drop across a 32-ohm resistor in a circuit powered by a 200-volt battery, which includes a parallel configuration of a 10-ohm and a 40-ohm resistor. The correct approach involves first determining the total resistance and current using Ohm's Law (V = IR). Additionally, the conversation addresses the heating of water using the 10-ohm resistor, requiring an understanding of the relationship between current, resistance, and heat transfer, specifically using the equation QV/t = I^2R.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Knowledge of series and parallel resistor configurations
  • Familiarity with thermal energy equations (QV/t = I^2R)
  • Basic principles of heat transfer in fluids
NEXT STEPS
  • Study the calculation of total resistance in parallel and series circuits
  • Learn about the temperature coefficient of resistivity for different materials
  • Explore the principles of thermal energy transfer in water heating
  • Investigate advanced applications of Ohm's Law in circuit analysis
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Students studying electrical engineering, physics enthusiasts, and anyone involved in circuit analysis or thermal dynamics in electrical systems.

scholio
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Homework Statement



a 200 volts battery is connected to three resistors as shown, see attachment.

a) what is the potential drop across the 32 ohm resistor?

b) the 10 ohm resistor is immersed in 60 grams of water at 20 deg celsius. how long much current flow for the water to reach 40deg celsius. assume that there is not heat loss from the water through the walls of the container.

Homework Equations



parallel resistors 1/R = 1/R_1 + 1/R_2

series resistors R = R_1 + R_2

potential V = IR where I is current, R is resistance

QV/t = I^2R where Q is charge, t is time

The Attempt at a Solution



part a. first i found the parallel resistor circuit between the 10 and 40 ohm resistor, getting 8 ohms,

then i connected the new 8ohm resistor to the 32 ohm and got a single resistor of 40 ohms.

i tried using V =IR to solve for potential, but it didn't work

QV/t = I^2R , couldn't use that and solve for V because i didn't have time

does the problem involve a temperature coefficient of resistivity of water to solve for part b?

any help, for either part, appreciated
 

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scholio said:
part a. first i found the parallel resistor circuit between the 10 and 40 ohm resistor, getting 8 ohms,
Correct.

then i connected the new 8ohm resistor to the 32 ohm and got a single resistor of 40 ohms.

i tried using V =IR to solve for potential, but it didn't work
You first have to determine the current. What is the current flowing through the 32 ohm resisitor (hint: it is a function of the total voltage and total resistance)? Then use V=IR to determine the voltage across the 32 ohm resistor.

QV/t = I^2R , couldn't use that and solve for V because i didn't have time

does the problem involve a temperature coefficient of resistivity of water to solve for part b?

any help, for either part, appreciated
Why use QV/t for power? Q/t = current = I so:

[tex]P = VI = I^2R[/tex]

AM
 

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