Kirchoff's Law - Wire in parallel with light bulb

AI Thread Summary
The discussion focuses on the effects of connecting a bare wire in parallel with a light bulb (L2) in a circuit containing three light bulbs with specified resistances. The equivalent resistance of the circuit becomes zero due to the wire, leading to the conclusion that the current will primarily flow through the wire, resulting in zero voltage across L2 and the third bulb (L3). Consequently, the brightness of L2 and L3 will decrease significantly, while L1 will maintain the full battery voltage. The analysis confirms that the power through L3 becomes negligible after the wire is connected. Understanding these principles illustrates the implications of Kirchhoff's Law in circuit behavior.
flannabhra
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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
upload_2015-5-31_2-34-15.png

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!
 
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flannabhra said:
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)?
 
Svein said:
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.
 
flannabhra said:
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...
 
It should be equal to the voltage of the battery. Thanks for your help.
 
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