# of elecron/sec in a 3 bulb circuit

[iknowless]

Homework Statement

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)

Homework 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

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

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!

 

[Borek]

Isn't it just about voltage, current and combining resistances?

 

[iknowless]

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.

 

[iknowless]

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

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

 

[asteriatic]

Based on the given information, we can calculate the electron current in circuit 1 as:

i_1 = nA_a u E_a = nA_b u E_b

Since A_b > A_a, we can rewrite this as:

i_1 = nA_b u (E_a / E_b) E_b

Now, from the given equations, we can see that E_a / E_b = E_c / (2E_b), so we can substitute this in:

i_1 = nA_b u (E_c / (2E_b)) E_b = nA_b u E_c / 2

Similarly, we can calculate the electron current in circuit 2 as:

i_2 = nA_a u E_c = nA_b u (E_c / 2)

Finally, in circuit 3, we have:

i_3 = nA_b u E_d

Since E_c = E_d, we can equate the electron currents in circuit 2 and 3:

nA_b u (E_c / 2) = nA_b u E_d

Solving for n, we get:

n = (2E_d) / (E_c u)

Therefore, the number of electrons per second flowing through the long bulb in circuit 3 is:

i_3 = nA_b u E_d = (2E_d) / (E_c u) * A_b u E_d = 2A_b (E_d)^2 / (E_c u)

Note: This answer assumes that the bulbs are connected in series, as stated in the problem. If the bulbs were connected in parallel, the number of electrons per second flowing through each bulb would be the same.