Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description here

The book said that the brightness of R1 would increase and R2 would decrease. I don't understand this at all.

Why does adding a wire from b to c change anything to R1?

Shouldn't charge still flow through R1 as they should and R2's brightness decrease because the charges at point b have an alternate route to travel to point 2?

share|cite|improve this question
up vote 2 down vote accepted

The voltage drop across R2 becomes 0, and the full voltage is applied across R1 instead.

share|cite|improve this answer
Why? I don't see that happening at all. I still see charges going through the lightbulb. Doesn't the current split at point b? – Hawk May 16 '12 at 3:32
Sure. But the resistance through the wire is much lower than the resistance through the bulb filament, causing the wire to shunt most of the current instead. – Ignacio Vazquez-Abrams May 16 '12 at 3:34
Is it because some dumb charges decides to go through the resistor, and the rest of the charges go to the alternative path after seeing the dumb charges struggle in the resistor? – Hawk May 16 '12 at 3:36
I don't know the exact particle mechanics of how parallel circuits work, but doing the math gives a lower resistance for it as compared to the constituent resistances. Since the current is constant through R1 and the new parallel circuit, and the resultant voltage drop must be proportional to the new resistance, there is effectively no voltage drop across the parallel circuit. – Ignacio Vazquez-Abrams May 16 '12 at 3:41
The energy in a circuit is not carried in the electrons (or rather, only a negligible fraction), the energy is carried in fields surrounding the wire. When you short-circuit a resistor, the field rearranges at nearly the speed of light to make the voltage at the two ends near zero, the reason is that the initial big voltage draws a large current through the small resistance, increasing the current fraction until the voltage at the end equilibrates at the Ohm's law value. – Ron Maimon May 16 '12 at 6:04

You ask: Shouldn't charge still flow through R1 as they should?

Well, yes, current still flows through R1 once the wire is put across b-c. But is it the same amount of current? What determines the amount of current flowing in the whole circuit? (there are two elements: do both of them stay the same when you add the wire?)

There are specific equations that tell you:

  1. the equivalent resistance of two resistances in parallel;
  2. the equivalent resistance of two resistances in series; and
  3. how much current flows in a circuit, given the voltage and the circuit resistance.

Start by writing all three down - add them to the question here. And now apply them to the circuit without the wire in place. And then apply them with the wire in place. This will explain why the brightness of each bulb changes.

If this only gets you part of the way, add your working to the question above, and I'll extend this answer.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.