Answer:
circuit sketched in first attached image.
Second attached image is for calculating the equivalent output resistance
Explanation:
For calculating the output voltage with regarding the first image.
![Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V](https://tex.z-dn.net/?f=Vout%20%3D%205%20%5Cfrac%7B2000%7D%7B5000%7D%5B%2F%5Btex%5D%3C%2Fp%3E%3Cp%3E%5Btex%5DVout%20%3D%205%20%5Cfrac%7B2000%7D%7B5000%7D%5C%5CVout%20%3D%205%20%5Cfrac%7B2%7D%7B5%7D%20%3D%202%20V)
For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.
so.
Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.
if the -5% is applied to both resistors the Voltage is still 5V because the quotient has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:
so.
Answer:
zero or 2π is maximum
Explanation:
Sine waves can be written
x₁ = A sin (kx -wt + φ₁)
x₂ = A sin (kx- wt + φ₂)
When the wave travels in the same direction
Xt = x₁ + x₂
Xt = A [sin (kx-wt + φ₁) + sin (kx-wt + φ₂)]
We are going to develop trigonometric functions, let's call
a = kx + wt
Xt = A [sin (a + φ₁) + sin (a + φ₂)
We develop breasts of double angles
sin (a + φ₁) = sin a cos φ₁ + sin φ₁ cos a
sin (a + φ₂) = sin a cos φ₂ + sin φ₂ cos a
Let's make the sum
sin (a + φ₁) + sin (a + φ₂) = sin a (cos φ₁ + cos φ₂) + cos a (sin φ₁ + sinφ₂)
to have a maximum of the sine function, the cosine of fi must be maximum
cos φ₁ + cos φ₂ = 1 +1 = 2
the possible values of each phase are
φ1 = 0, π, 2π
φ2 = 0, π, 2π,
so that the phase difference of being zero or 2π is maximum
Answer:
57.94°
Explanation:
we know that the expression of flux
where Ф= flux
E= electric field
S= surface area
θ = angle between the direction of electric field and normal to the surface.
we have Given Ф= 78 
E=
S=
= 
=0.5306
θ=57.94°
Answer:
Explanation:
b ) First is concave lens with focal length f₁ = - 12 cm .
object distance u = - 20 cm .
Lens formula
1 / v - 1 / u = 1 / f
1 / v + 1 / 20 = -1 / 12
1 / v = - 1 / 20 -1 / 12
= - .05 - .08333
= - .13333
v = - 1 / .13333
= - 7.5 cm
first image is formed before the first lens on the side of object.
This will become object for second lens
distance from second lens = 7.5 + 9 = 16.5 cm
c )
For second lens
object distance u = - 16.5 cm
focal length f₂ = + 12 cm ( lens is convex )
image distance = v
lens formula ,
1 / v - 1 / u = 1 / f₂
1 / v + 1 / 16.5 = 1 / 12
1 / v = 1 / 12 - 1 / 16.5
= .08333- .0606
= .02273
v = 1 / .02273
= 44 cm ( approx )
It will be formed on the other side of convex lens
distance from first lens
= 44 + 9 = 53 cm .
magnification by first lens = v / u
= -7.5 / -20 = .375 .
magnification by second lens = v / u
= 44 / - 16.5
= - 2.67
d )
total magnification
= .375 x - 2.67
= - 1.00125
height of final image
= 2.50 mm x 1.00125
= 2.503mm
e )
The final image will be inverted with respect to object because total magnification is negative .
Answer: -2.5
Explanation:
1/2(-5)= -2.5
-2.5(1)= -2.5
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