Solving 10 Ohm Resistor Problem: Min 3 in Series/Parallel

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SUMMARY

The minimum number of 10 ohm resistors required to achieve a total resistance of 10 ohms capable of dissipating at least 5 watts is nine. This configuration consists of three sets of three 10 ohm resistors connected in series, which results in a total resistance of 30 ohms per set. These sets are then connected in parallel, allowing for a total power dissipation of 9 watts. Understanding the calculations involving Ohm's law and the formulas for equivalent resistance in series and parallel arrangements is essential for solving such circuit problems.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Knowledge of series and parallel resistor configurations
  • Familiarity with power dissipation calculations (P=RI²)
  • Ability to solve systems of inequalities in circuit analysis
NEXT STEPS
  • Study the formulas for equivalent resistance in series and parallel circuits
  • Learn how to calculate power dissipation in resistors
  • Explore advanced circuit analysis techniques, including mesh and nodal analysis
  • Practice solving complex resistor network problems
USEFUL FOR

Electrical engineering students, circuit designers, and hobbyists interested in understanding resistor configurations and power management in electronic circuits.

Berg
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You are given a number of 10 ohm resistors, each capable of dissopating only 1 w without being destroyed. What is the minimum number of such resistors that you need to combine in series or parrallel to make a 10 ohm resistance that is capable of dissipating at least 5w? Hmm i know the answer is 3 series of 3 in parrallel but i don't know how to work to achieve this answer.

Also when doing circuit problems do u find the current at each resistor? or each branch?
 
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Each branch from a circuit has a current (or intensity,the old name).So that should be very clear.As for your problem,it's not that simple.U must take into acount that
[tex[ P=RI^{2} [/tex] and a bunch of inequalities forthe various powers dissipated in resistors.Plus u'll need to know the formulas giving the equivalent resistance for a seris arrangement & for a parallel one,respectively.U should (hopefully) be able to write down a system of inequalities which MUST BE SOLVED INTO [tex]\mathbb{N}[/tex].

Good luck!

Daniel.
 


To solve this problem, we can use a combination of series and parallel connections for the resistors. First, we can connect three 10 ohm resistors in series, resulting in a total resistance of 30 ohms. This will also limit the power dissipation to 1 watt for each resistor, ensuring they are not destroyed.

Next, we can connect three sets of these series-connected resistors in parallel to each other. This will result in a total resistance of 10 ohms, as each set of series-connected resistors will have a total resistance of 30 ohms. This parallel connection will also increase the power dissipation capability to 3 watts for each set of resistors, giving us a total power dissipation of 9 watts (3 watts x 3 sets).

Therefore, by combining three sets of three 10 ohm resistors in series/parallel, we can achieve a 10 ohm resistance with a power dissipation capability of 9 watts, meeting the requirements of the problem.

When solving circuit problems, it is important to consider the current at each resistor and branch. This will help us determine the power dissipation and voltage drop at each point in the circuit. By using Ohm's law (V=IR), we can calculate the current at each resistor by dividing the voltage drop across it by its resistance. Similarly, for each branch, we can calculate the total current by adding the currents at each resistor in that branch. This information is crucial in understanding the behavior of the circuit and making accurate calculations.
 

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