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iknowless

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## 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

[tex]i_1=nA_auE_a=n2A_buE_b[/tex]

[tex]i_2=nA_auE_c=2.5e18[/tex]

[tex]i_3=nA_buE_d=5e18[/tex]

[tex]2emf=l(E_a+2E_b)=E_c l=E_d l[/tex]

## The Attempt at a Solution

note: [itex]A_b>A_a[/itex]

[tex]E_c=E_d[/tex]

I am assuming that all the filaments have a length 'l.'

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

From [itex]i_1[/itex], [itex]A_a E_a=2A_b E_b[/itex]

From [itex]\frac{i_1}{i_2}[/itex], [itex]E_a i_2 =2 E_b i_3[/itex]

Also, [itex]i_1=\frac{E_a i_2}{E_c}=\frac{2 E_b i_3}{E_c}[/itex]

And [itex]\frac{i_2}{i_3} = \frac{A_a}{A_b}[/itex]

[tex]E_a+2E_b=2E_b+E_b \frac{i_3}{i_2}=E_c[/tex]

Please just give me a nudge in the right direction. Thanks!

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