# DC circuit problem

1. Jan 23, 2009

### Elbobo

1. The problem statement, all variables and given/known data
Two identical light bulbs A and B are connected in series to a constant voltage source.
Suppose a wire is connected across bulb B as shown (basically, the wire is connected to both ends of bulb B; light bulb A is closer to the positive side of the cell).

Bulb A:

1. will burn twice as brightly as before.
2. will burn half as brightly as before.
3. will burn as brightly as before.
4. will burn nearly four times as brightly as
before.
5. will go out

2. Relevant equations
V= IR
P = VI

3. The attempt at a solution
I figured the wire's resistance would almost nothing, so I just ignored it and that would mean that all the current should just flow through bulb B as it would without the wire. So, yeah.... I'm wrong.

(i.e., not choice 3)

2. Jan 23, 2009

### rl.bhat

When the bulb is short circuited by a wire, the total resistance of the circuit reduces to half. Brightness of the bulb depends on the power consumption. Since apply voltage remains constant and P = V^2/R , decide the correct choice.

3. Jan 23, 2009

### Elbobo

Is that just a general rule of resistors being short circuited? My teacher and my textbook haven't even mentioned it.

P(old) = (V^2)/(2R)

P(new) = V^2 / (Rnew)

Rnew = R (since you said it should be half)

P(new) = V^2 / R

It should be twice as bright as before, but that answer is incorrect.

Last edited: Jan 23, 2009
4. Jan 23, 2009

### rl.bhat

Bulb B has a higher resistance than the wire. So most of the current will flow through the wire than the bulb B. The bulb B will off.

5. Jan 23, 2009

### Elbobo

Oh, I understand the halving. But still, why does bulb A not shine twice as brightly?

6. Jan 23, 2009

### chrisk

An electric current will seek the path of least resistance. Since Bulb B was connected in parallel with a wire with very little resistance, the bulk of the current will pass through the wire and Bulb B will not light. This also means the resistance of the circuit has decreased, and in a series circuit the current will increase. Power = i2R where R is the resistance of Bulb A. This information is sufficient to answer the question.

7. Jan 23, 2009

### rl.bhat

Oh, I understand the halving. But still, why does bulb A not shine twice as brightly?

It depends on the rating of the bulb A. If the new current is more than it can withstand, the bulb will go out.

8. Jan 23, 2009

### Elbobo

OK, I finally got the right answer, but that was by letting the power of Bulb A equal (I^2)*2R. I thought power at an individual resistor on a series would only rely on that resistor's resistance, not the entire circuit's resistance.

So the latter's how you would calculate power in all DC circuit cases?