Answer:
a.) I = 7.8 × 10^-4 A
b.) V(20) = 9.3 × 10^-43 V
Explanation:
Given that the
R1 = 20 kΩ,
R2 = 12 kΩ,
C = 10 µ F, and
ε = 25 V.
R1 and R2 are in series with each other.
Let us first find the equivalent resistance R
R = R1 + R2
R = 20 + 12 = 32 kΩ
At t = 0, V = 25v
From ohms law, V = IR
Make current I the subject of formula
I = V/R
I = 25/32 × 10^3
I = 7.8 × 10^-4 A
b.) The voltage across R1 after a long time can be achieved by using the formula
V(t) = Voe^- (t/RC)
V(t) = 25e^- t/20000 × 10×10^-6
V(t) = 25e^- t/0.2
After a very long time. Let assume t = 20s. Then
V(20) = 25e^- 20/0.2
V(20) = 25e^-100
V(20) = 25 × 3.72 × 10^-44
V(20) = 9.3 × 10^-43 V
Answer:
Q(h=200)=0.35W
Q(h=3000)=5.25W
Explanation:
first part h=200W/Km^2
we must use the convection heat transfer equation for the chip
Q=hA(Ts-T∞)
h=
convective coefficient=200W/m2 K
A=Base*Leght=5mmx5mm=25mm^2
Ts=temperature of the chip=85C
T∞=temperature of coolant=15C
Q=200x2.5x10^-5(85-15)=0.35W
Second part h=3000W/Km^2
Q=3000x2.5x10^-5(85-15)=5.25W
You could search it sorry I didn’t fiv you the answer
Answer: the final temperature of the steam 581.5 °C
Explanation:
Given that;
P₁ = 11 MPa
T₁ = 600°C
exit at; P₂= 5.5 MPa
Now from superheated steam table( p=11 MPa, T=600°C)
h₁ = 3615 kJ/kg
h₁ = h₂ ( by throttling process and adiabatic isentholpic )
from saturated steam table at; ( h= 3615 kJ/kg, P= 5.5 MPa )
Temperature = 581.5 °C
Therefore the final temperature of the steam 581.5 °C
Answer: Context
Explanation: It is always very important for an engineer to keep the context of his/her expirament in mind.
