Determining voltage across resistor and current direction

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SUMMARY

The discussion focuses on determining the voltage across a resistor and the direction of current flow in an electrical circuit involving a resistor with a resistance of 8 ohms and a power dissipation of 11 watts. Using the formula V = IR, the voltage across the resistor is calculated to be 9.33 volts. The current entering the network from the resistor is determined to be -1.17 A, indicating that the current is leaving the network, which is consistent with the passive nature of the resistor.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Knowledge of power calculations in electrical circuits (P = I²R)
  • Familiarity with passive components in electrical engineering
  • Basic concepts of current direction and sign conventions
NEXT STEPS
  • Research the implications of passive components in circuit analysis
  • Learn about sign conventions in electrical engineering
  • Explore advanced power calculations in AC circuits
  • Study the effects of resistance on current flow in various circuit configurations
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Electrical engineering students, circuit designers, and anyone studying the principles of current flow and power dissipation in resistive circuits.

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Homework Statement



Resistor conected to unknown network N. Resistor immersed in isolated water bath. Resistor resistance R = 8 ohms. It is determined the power dissipated in resistor is 11 W

a) Find the voltage across the resistor.

b) What is the current i entering the network N from the resistor?
[/B]
upload_2015-1-15_17-4-3.png


Homework Equations



V=IR

The Attempt at a Solution



Ignore part a). I understand my mistake for that one.

For part b) though I am trying to get a better intuition for what in this problem would tell us that current is actually leaving the network as opposed to entering it. What is the physical intuition for why that is the case and what is the matematical reason part b) has a negative -1.17 A?
 
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The resistor is a passive element, it does not source current. So we know the current through the resistor is coming out of the network, and it's 1.17A.

So, 1.17A out of the network can be expressed as negative current INTO the network. The question asks for the current entering the network from the resistor, this must therefore be negative 1.17A, i.e., -1.17A. It's just the way it's done to stay mathematically correct at all times.

Power is I2.R so any negative sign disappears in the power calculation.
 

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