Sorry I could not get the image to upload please help though thank you! Two iden

Question

Sorry I could not get the image to upload please help though thank you!

Two identical light bulbs A and B are con- nected in series to a constant voltage source. Suppose a wire is connected across bulb B as shown.

Bulb A

1. will burn half as brightly as before.

2. will go out.

3. will burn as brightly as before.

4. will burn nearly four times as brightly as before.

5. will burn twice as brightly as before.

and bulb B
1. will burn half as brightly as before. 2. will go out.

3. will burn nearly four times as brightly as before.

4. will burn as brightly as before.
5. will burn twice as brightly as before.

Explanation / Answer

4. will burn nearly four times as brightly as before.

A light bulb will have a certain resistance R. The light it emits will be proportional to the energy dissipated by this resistance P = i2 R

So when two identical light bulbs are connected in series to a constant voltage V, the current passing through the circuit is i = V / 2R and the power dissipated in light bulb A is:

P1 = (V2 / 4R2) R = V2 / 4R

Now when a wire is connected across light bulb B, that means that the resistance is bypassed by a wire in parallel to the bulb (i.e. the bulb is shorted out by a jumper).

The resistance of bulb B becomes zero, and the current in the circuit is i = V/R

This current is twice as high as before, but the power dissipated by light bulb A will be four times more:

P2 = V2 / R = 4 P1

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