When the input voltages of a difference amplifier are 5.1 v and 6.4 v, the output voltage is 64.7 v. The inputs are changed to 4
1 answer:
Answer:
a) Differential mode gain = 48
b) Common mode gain = 0.4
c) CMRR = 120
Explanation:
The output of a difference amplifier is related to the input by the equation:
When V₁ = 6.4 V, V₂ = 5.1 V and V₀ = 64.7 V, the equation becomes
6.4 A₁ + 5.1 A₂ = 64.7.....................(1)
When V₁ = 5.6 V, V₂ = 4.9 V and V₀ = 35.7 V, the equation becomes
5.6 A₁ + 4.9 A₂ = 35.7.....................(2)
Multiply equation (1) by 5.6 and (2) by 6.4
35.84 A₁ + 28.56A₂ = 362.32.....................(3)
35.84 A₁ + 31.36 A₂ = 228.48....................................(4)
Subtract equation (3) from (4)
2.8 A₂ = -133.84
A₂ = -133.84/2.8
A₂ = -47.8
Put the value of A₂ into equation (1)
6.4 A₁ + 5.1 (-47.8) = 64.7
6.4 A₁ = 64.7 + 243.78
A₁ = 308.48/6.4
A₁ = 48.2
a) Common mode gain = A₁ + A₂ = 48.2 + (-47.8)
Common mode gain = 0.4
b) Differential mode gain = (A₁ -A₂)/2
Differential mode gain = (48.2 - (-47.8))/2
Differential mode gain = 96/2
Differential mode gain = 48
c) Common Mode Rejection Ratio (CMRR)
You might be interested in
Answer:
magnitude: 21.6; direction: 33.7 degrees
Explanation:
When we multiply a vector by a scalar, we have to multiply each component of the vector by the scalar number. In this case, we have
vector: (-3,-2)
Scalar: -6
so the vector multiplied by the scalar will have components
The magnitude is given by Pythagorean's theorem:
and the direction is given by the arctan of the ratio between the y-component and the x-component:
Answer:
v_{1fy} = - 0.4549 m / s
Explanation:
This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved
initial. Before the crash
p₀ = m v₁₀
final. After the crash
= m
+ m v_{2f}
Recall that velocities are a vector so it has x and y components
p₀ = p_{f}
we write this equation for each axis
X axis
m v₁₀ = m v_{1fx} + m v_{2fx}
Y Axis
0 = -m v_{1fy} + m v_{2fy}
the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components
sin 23.3 = v_{2fy} / v_{2f}
cos 23.3 = v_{2fx} / v_{2f}
v_{2fy} = v_{2f} sin 23.3
v_{2fx} = v_{2f} cos 23.3
we substitute in the momentum conservation equation
m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3
0 = - m v_{1f} sin θ + m v_{2f} sin 23.3
1.83 = v_{1f} cos θ + 1.15 cos 23.3
0 = - v_{1f} sin θ + 1.15 sin 23.3
1.83 = v_{1f} cos θ + 1.0562
0 = - v_{1f} sin θ + 0.4549
v_{1f} sin θ = 0.4549
v_{1f} cos θ = -0.7738
we divide these two equations
tan θ = - 0.5878
θ = tan-1 (-0.5878)
θ = -30.45º
we substitute in one of the two and find the final velocity of the incident ball
v_{1f} cos (-30.45) = - 0.7738
v_{1f} = -0.7738 / cos 30.45
v_{1f} = -0.8976 m / s
the component and this speed is
v_{1fy} = v1f sin θ
v_{1fy} = 0.8976 sin (30.45)
v_{1fy} = - 0.4549 m / s