Given V1 = 8 vpp, R1 = 6 kΩ, V2 = 5 vpp, R2 = 3 kΩ, V3 = 10 vpp, R3 = 3 kΩ, and Rf = 14 kΩ, find Vout. Vout = vpp. (Round your a
nswer to 2 decimal places.) In phasor notation, Vout = at an angle of o vpp.
1 answer:
Answer:
Vout= 93.3V
Explanation:
For this question, consider circuit in the attachment 1.
This is the circuit of an inverting amplifier. In an inverting amplifier
Vout/Vin= -Rf/Rin
To calculate the Vout, we must find Rin and Vin. For this we must solve the input circuit (attachment 2) using Thevinine theorem. Thevnine theorem states that all voltage sources in a circuit can be replaced by an equivalent voltage source Veq and and all resistances can be replaced by an equivalent resistance Req. To find out Req all voltage sources must be short circuited (attachment 3)
1/Req= 1/R1+1/R2+1/R3
1/Req=1/6+1/3+1/3
Req=6/5
To find out Veq consider circuit in attachment 4. We will solve this circuit using nodal analysis. In nodal analysis, we use the concept that sum of currents entering a node is equal to the sum of currents leaving a node. So,
I1= I2+I3
(10-Veq)/6= (Veq-5)/3+(Veq-10)/3
Veq=8V
Now the input circuit can be simplified as shown in attachment 5. Solve for Vout using equation
Vout/Veq= -Rf/Req
Vout/8= -14/(6/5)
Vout= - 93.3
It is at an angle of 180° from Veq
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Answer:
a) E = b) i.
Explanation:
Assuming no other forces acting on the electron, as the field is pushing the electron back till it comes to rest momentarily, applying the work-energy theorem, the loss in kinetic energy, must be equal to the work done by the field on the electron.
The force that does work on the electron is the one due to the electric field E, and by definition of electric field, can be expressed as follows:
W = q*E = -e*E (1)
This work must be equal to the change in Kinetic Energy:
ΔK = Kf -K₀ = 0 - K = -K (2)
From (1) and (2), and solving for E, we have:
E =
b) As the electric field (by convention) has the direction that it would have a positive test charge, the electrons (being negative charges),when starting from rest),have the direction opposite to the field.
In this case, as the electron has an initial velocity, and the field is opposing to the electron movement, we conclude that the electric field is in the direction of the electron´s motion, due to the electron slows down, and then comes momentarily to an stop, before changing direction and move opposite to the field as it is its natural behavior.