The circuit shown in the diagram is assembled from two 1.5 V batteries, thick co
ID: 1606846 • Letter: T
Question
The circuit shown in the diagram is assembled from two 1.5 V batteries, thick copper connecting wires, and two conducting rods. Rod 1 has a cross sectional area of 2.0 times 10^-6 m^2 and a length of 7 cm. Rod 2 has a cross sectional area of 1.5 times 10^-7 m^2 and is 16 cm long. The material of which both rods are made has 3 times 10^25 mobile electrons per cubic meter, and an electron mobility of 3 times 10^-4 m/s/N/C (a) What is the magnitude of the electric field at location A in rod 1? Start from fundamental rules and (b) What is the magnitude of the drift speed v_1 of a mobile electron at location A in rod 1? (c) What is the conventional current passing through rod 1? (d) What is the resistance of rod 1 in ohms?Explanation / Answer
(a)
The loop equation is (E = EMF of battery):
+2E E1L1 E2L2 = 0
= 1.5 V
L1 =7 x10^-2 m
A1 = 2x10^-6 m2
A2 = 1.5X10^-7 m2
L2 =16x x10^-2 m
N= 3x10^25
= 3x 10^-4
where we are ignoring the small voltage drop in the thick copper wires.
The current equation is: i1 = i2 = N1A1u1E1 = N2A2u2E2, so
N1 = N2; u2 = u1 so
A1 E1 = A2 E2
E1 = 0.075E2
Putting everything together,
2 = E1xL1 +E2 XL2
6 = (0.075E2)x7 x10^-2 + E2x16x x10^-2
6 =5.25x 10^-3 E2 + E2x16x x10^-2
6 = E2(0.16)
E2 = 37.49 V/m
E1 = 2.81 V/m
(b)
u = E
where u = drift velocity; = mobility; E = electric field
u1= 3x 10^-4 x 2.81
u1 = 8.43 x 10^-4 m/s
(C)
I1 = A1 NE
=2x10^-6 x 3x10^25 x2.81x 3x 10^-4
=5 X10^16 electrons / seconds
= 8 mill ampere
d)
R = dV/ q(i)
= 3 /(1.6x10^-19 C x 5 x10^16 e/s)
=375
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