Adding a wrench to a series circuit...?

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

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SammyS

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.

ehild

the wrench is parallel with R2.
Correct, for the emf of the battery.

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.

mirandab17

I don't know how to go about finding the current through the wrench though... especially if it has negligible resistance.

mirandab17

oooh okay let me do that then...

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...

ehild

Excellent!

ehild

mirandab17

Thanks! Finally get it, awesome.