Three identical low-resistance (round) bulbs are in series as shown in the figur

Question

Three identical low-resistance (round) bulbs are in series as shown in the figure. Thick copper wires connect the bulbs. In the steady state, 2 1017 electrons leave the battery at location A every second.

How many electrons enter the second bulb at location D every second?

Next the middle bulb (at DE) is replaced by a wire, as shown in the figure:

Now how many electrons leave the batteries at location A every second? Do not use ohms or series-resistance formulas in your explanation, unless you can show in detail how these concepts follow from the microscopic analysis introduced in this chapter.

Explanation / Answer

A)
In series same current flows through all bulb.
so the answer will be 2*10^17 electron per second at D

B)
Now resistance has become 2/3 as before.
So number of electron will become 3/2 as before
So, number of electron flowing (3/2)*2*10^17 = 3*10^17 electrons

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