Resistors of 150 Ω and 100 Ω are connected in parallel. What is their equivalent resistance?
1 answer:
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Answer:
Explanation:
Given data:
initial construction co = 0.286 wt %
concentration at surface position cs = 0 wt %
carbon concentration cx = 0.215 wt%
time = 7 hr
for 0.225% carbon concentration following formula is used
where, erf stand for error function
from the table erf(Z) value = 0.751 lie between (z) = 0.80 and z = 0.85 so by inteerpolation we have z = 0.815
from given table
x = 0.002395 mm
a) Potential in A: -2700 V
b) Potential difference: -26,800 V
c) Work:
Explanation:
a)
The electric potential at a distance r from a single-point charge is given by:
where
is the Coulomb's constant
q is the charge
r is the distance from the charge
In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.
Charge 1 is
and is located at the origin (x=0, y=0)
Charge 2 is
and is located at (x=0, y = 0.40 m)
Point A is located at (x = 0.40 m, y = 0)
The distance of point A from charge 1 is
So the potential due to charge 2 is
The distance of point A from charge 2 is
So the potential due to charge 1 is
Therefore, the net potential at point A is
b)
Here we have to calculate the net potential at point B, located at
(x = 0.40 m, y = 0.30 m)
The distance of charge 1 from point B is
So the potential due to charge 1 at point B is
The distance of charge 2 from point B is
So the potential due to charge 2 at point B is
Therefore, the net potential at point B is
So the potential difference is
c)
The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by
where
q is the charge of the particle
is the potential difference
In this problem, we have:
is the charge of the electron
is the potential difference
Therefore, the work required on the electron is