Magnitudes of the currents are i1= 0.00104A , i2= 0.003641A , i3= 0.00508A.
Explanation:
Using superposition theorem,
remove the E1 voltage supply source and calculate resistance across it.
From the circuit given, as the resistances R1 and R2 are parallel
using the formula R1//R2=(R1.R2/(R1+R3)
V1= ((r1||r2)/(r1+r2||r3))*E1
v1 = (((1kΩ*680Ω)/(1kΩ+680Ω))/(3.3kΩ +((1kΩ*680Ω)/(1kΩ+680Ω)))*10v
v1= 2.3v
v2 = ((r1||r2)/(r1+r2||r3))*E2
v2 = (((1kΩ*680Ω)/(1kΩ+680Ω))/(3.3kΩ +((1kΩ*680Ω)/(1kΩ+680Ω)))*5v
v2= 1.161v
V = V1+V2
=> 2.3 + 1.161
=> 3.461v
Magnitudes of the currents can be found by i=v/r
i1 = v/r1
=> 3.461/3.3kΩ
=>0.00104A
i2=2.89/1kΩ
=>0.003461A
i3=2.89/680Ω
=> 0.00508A.
Therefore the magnitudes of the currents are i1, i2, and i3.
The three exposure techniques in photolithography are:
- Contact
- Proximity
- Projection
Alternatives to photolithography in IC processing include;
- X-ray
- UV
- Ion, and
- Electron lithography
<h3>
What is Photolithography?</h3>
Photolithography is a term in integrated circuit development that describes the patterned films that are formed when a beam of light falls on a substance.
This phenomenon protects the surface of sensitive materials such as glass during some operations like etching. UV and X-rays can be used for this purpose.
Learn more about photolithography here:
brainly.com/question/13650094
#SPJ11
Answer:no fire can be used on fire
Explanation:
because no fire can be used on fire
Answer:
9V
Explanation:
I'm not sure if this is correrct, but i assume you just do a loop in the path of D1 to the resistor to D3. Using kirchoff voltage law, they should sum up to 0. The forward voltage drop of each diode is already given so just sub those value i and solve for Vr (voltage drop over resistor).
10 - Vd1 - Vr - Vd3 = 0
10 - .3 - Vr - .7 = 0
Vr = 10 - .3 - .7 = 9V