Question 1 (20 point) Bulb 1 Bulb 2 EE D In the circuit shown here, bulbs 1 and

Question

Question 1 (20 point) Bulb 1 Bulb 2 EE D In the circuit shown here, bulbs 1 and 2 are identical in mechanical construction (both filaments have the same length of 4 mm and the same cross sectional area 2.5e-8 m but are made of the different metals. The electron mobility in the metal used in the bulb 2 is four times as large as electron mobility in the metal used in the bulb 1,but both metals have 6.3e28 mobile wires. The electrons/m'. The two bulbs are connected in series to two batteries with thick copper emf of each battery is 1.5 volts. each bulb. Start from A) Calculate the magnitude of the electric fields Ei and Ez inside fundamental principles and show all your steps. 8 pts) E, s A Seema Go e e o.odu) 4,31 e Loge B) If the electron mobility of the met your work. (4 pts) second pass through bulb 1? Show all your steps in 11e

Explanation / Answer

a)
Take E1 be the electric field across bulb1 and E2 be the electric field across bulb2.
Voltage across the bulbs = 2 x 1.5 = 3 V
V = E1d + E2d    (d being the length of the filament)
3 = (E1 + E2) x 4 x 10-3
E1 + E2 = 750    ...(1)

Same current flows through both the bulbs since they are connected in series.
Drift velocity, vd = I/nAe
n is the charge density, A is the area and e is the charge of electron, since both the bulbs have these quantities same, they have equal drift velocities.
Electric field = vd/
Where is the electron mobility
Given that 41 = 2
E1 = vd/1, E2 = vd/41
E1/E2 = 4, E1 = 4E2      ...(2)

Substituting in (1)
4E2 + E2 = 750
E2 = 150 V/m
E1 = 4E2 = 600 V/m.

b)
Drift velocity, vd = E11
= 600 x 8 x 10-5 = 4.8 x 10-2 m/s.

Electrons per second = Current, I = nAevd
= (6.3 x 1028) x (2.5 x 10-8) x (1.6 x 10-19 ) x (4.8 x 10-2)
= 12.096 A.

c)
Potential difference in the first bulb
V1 = E1 x d
= 600 x 4 x 10-3
= 2.4 V, Its direction is from C to B.

d)
Resistance, R1 = V1/I
= 2.4 / 12.096
= 0.198 Ohm

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