## # of elecron/sec in a 3 bulb circuit

1. The problem statement, all variables and given/known data

When a single round bulb of a particular kind and two batteries are connected in series, 5* 10^18 electrons pass through the bulb every second. When a single long bulb of a particular kind and two batteries are connected in series, only 2.5*10^18 electrons pass through the bulb every second, because the filament has a smaller cross section.

Figure 18.68
In the circuit shown in Figure 18.68, how many electrons per second flow through the long bulb?

E is the electric field in the filament
A is the cross-sectional area of the filament
i is electron current
l is the length of the filament
n is the # of electrons per unit volume (unknown)
u is the electron mobility (unknown)

2. Relevant equations

Circuit 1 is the main circuit, with long bulb A, and two identical round bulbs, both labeled B.
Circuit 2 is the circuit with 1 long bulb, C.
Circuit 3 is the circuit with 1 round bulb, D

$$i_1=nA_auE_a=n2A_buE_b$$
$$i_2=nA_auE_c=2.5e18$$
$$i_3=nA_buE_d=5e18$$
$$2emf=l(E_a+2E_b)=E_c l=E_d l$$

3. The attempt at a solution

note: $A_b>A_a$
$$E_c=E_d$$
I am assuming that all the filaments have a length 'l.'

I have created fractions of current to produce some hopefully useful formulae.

From $i_1$, $A_a E_a=2A_b E_b$

From $\frac{i_1}{i_2}$, $E_a i_2 =2 E_b i_3$

Also, $i_1=\frac{E_a i_2}{E_c}=\frac{2 E_b i_3}{E_c}$

And $\frac{i_2}{i_3} = \frac{A_a}{A_b}$

$$E_a+2E_b=2E_b+E_b \frac{i_3}{i_2}=E_c$$

Please just give me a nudge in the right direction. Thanks!!!
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 We haven't talked about resistance or Ohm's law yet. We can only use Kirchhoff's node rule, "In the steady state, the electron current entering a node in a circuit is equal to the electron current leaving the node." We can also use "i=nAuE" and the Loop Rule, where voltage around a closed path is zero in a circuit. Is there an easier way to do it with resistance? Please help, but I would also like to know how to do it without Ohm's Law.

## # of elecron/sec in a 3 bulb circuit

Wow, how did I not see it? In formula 4

 Quote by iknowless And $\frac{i_2}{i_3} = \frac{A_a}{A_b}$
Then $A_b=2A_a$ !

So now formula 1 can be written as $E_a=4E_b$

There is an error in the 4th relevant equation. It should be $2emf=l(E_a+E_b)=lE_c$ because the closed path only travels through 1 round bulb, not 2.

and the rest is algebra

 Tags bulb, circuit, electron