Kirchoff's Law - Wire in parallel with light bulb

[flannabhra]

Homework Statement

If you measure the resistance of the three lightbulbs (independently) and get the following values. What is the equivalent total resistance of the circuit, and it’s uncertainty.
$$R_1 = (50 \pm 7) \Omega, \ \ R_2 = (720 \pm 30) \Omega, \ \ R_3 = 140 \pm 20) \Omega$$

Now you connect a bare wire (with negligible internal resistance) between to the left and right of L2 respectively in the circuit diagram. What will happen to the brightness of each bulb (i.e. will it increase, decrease, or remain the same)? Justify your results by finding the power of each light bulb in terms of ∆Vbat.

Homework Equations

$$\frac{1}{R_{eq}} = \frac{1}{R_{L2}} + \frac{1}{0}$$

The Attempt at a Solution

$$\frac{1}{R_{eq}} = \frac{1}{720 \ \Omega} + \frac{1}{0}$$
$$R_{eq} = 0 \Omega$$

Knowing that the resistance in the top half of the parallel component of this circuit is now 0, does that imply that the current, I1, into the parallel component is now equivalent to I2, the current into L2 and the wire? I.E. is there any power going through L3 after the wire is connected?

Thank you in advance!

 

[Svein]

Knowing that the resistance in the top half of the parallel component of this circuit is now 0, does that imply that the current, I1, into the parallel component is now equivalent to I2, the current into L2 and the wire? I.E. is there any power going through L3 after the wire is connected?

Well, what do you think? What is the voltage across L2 (and L3)?

 

[flannabhra]

Well, what do you think? What is the voltage across L2 (and L3)?

I think that all (minus a negligible amount) the current will flow through the wire, making the Voltage drop across L2 and L3 zero.

 

[Svein]

I think that all (minus a negligible amount) the current will flow through the wire, making the Voltage drop across L2 and L3 zero.

Yes. Therefore, the voltage across L1 is...

 

[flannabhra]

It should be equal to the voltage of the battery. Thanks for your help.