Adding a wrench to a series circuit ?

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Homework Help Overview

The discussion revolves around the configuration of a circuit when a wrench is added, specifically whether this addition creates a parallel circuit. The subject area involves circuit analysis, particularly focusing on Ohm's Law and Kirchhoff's Laws.

Discussion Character

  • Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of adding a wrench to the circuit, questioning whether it alters the circuit configuration to parallel. They discuss the initial voltage calculation and the current through the resistors, with some uncertainty about how the wrench's resistance affects the overall current distribution.

Discussion Status

There is an ongoing exploration of the circuit behavior after the wrench is added, with some participants providing calculations and insights into the voltage and current through the resistors. While some guidance has been offered regarding the application of Kirchhoff's Laws, there is no explicit consensus on the final interpretation of the circuit's behavior.

Contextual Notes

Participants are working under the assumption that the wrench has negligible resistance, which raises questions about how to accurately determine the current through it. The discussion reflects a mix of assumptions and interpretations regarding the circuit's configuration and behavior.

mirandab17
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If someone could also clarify how to go about this one.

Does the wrench added make it a parallel circuit?
I found the voltage of the initial circuit by simply doing V = IR = (3)(10) = 30 V.

The current flowing through the 4 ohm resistor must be the same as before... so the current for that is still 3.0 A?

Help please :s
 

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mirandab17 said:
If someone could also clarify how to go about this one.

Does the wrench added make it a parallel circuit?
I found the voltage of the initial circuit by simply doing V = IR = (3)(10) = 30 V.

The current flowing through the 4 ohm resistor must be the same as before... so the current for that is still 3.0 A?

Help please :s
attachment.php?attachmentid=44481&d=1330405295.png


Well, in a sense the wrench makes it a parallel circuit, with the wrench being in parallel with R2. But the resistance of the wrench is small. (I assume that it's much smaller than the resistance of either resistor.) The effective resistance of two resistors in parallel is less than the resistance of either one.
 
mirandab17 said:
If someone could also clarify how to go about this one.

Does the wrench added make it a parallel circuit?
the wrench is parallel with R2.
mirandab17 said:
I found the voltage of the initial circuit by simply doing V = IR = (3)(10) = 30 V.

Correct, for the emf of the battery.

mirandab17 said:
The current flowing through the 4 ohm resistor must be the same as before... so the current for that is still 3.0 A?

NO! It is the emf that is the same as before, 30 V. The currents will change. What is the voltage across R2 if 1 A current flows through it? What is then the voltage across R1? And what current flows through it? And then apply Kirchhoff's current Law.
 
I don't know how to go about finding the current through the wrench though... especially if it has negligible resistance.
 
oooh okay let me do that then...
 
Voltage across R2
V = IR = (1)(6) = 6 V

Is the voltage across R1 the EMF minus the voltage parallel?
So voltage across R1 = 30 - 6 = 24 V?
Current through R1 would be I=V/R = 24/4 = 6 A.

Oh! And then if there's only 1 A through R2 then there must be 5 A through the wrench due to Kirchoff's current law...
 
mirandab17 said:
Voltage across R2
V = IR = (1)(6) = 6 V

Is the voltage across R1 the EMF minus the voltage parallel?
So voltage across R1 = 30 - 6 = 24 V?
Current through R1 would be I=V/R = 24/4 = 6 A.

Oh! And then if there's only 1 A through R2 then there must be 5 A through the wrench due to Kirchoff's current law...

Excellent!:smile:

ehild
 
Thanks! Finally get it, awesome.
 

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