## Adding a wrench to a series circuit...?

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?

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 Quote by mirandab17 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

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.

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 Quote by mirandab17 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.
 Quote by mirandab17 I found the voltage of the initial circuit by simply doing V = IR = (3)(10) = 30 V.
Correct, for the emf of the battery.

 Quote by mirandab17 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.

## Adding a wrench to a series circuit...?

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...

Recognitions:
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 Quote by mirandab17 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!

ehild
 Thanks! Finally get it, awesome.

 Tags resistance, resistors

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