Pick one product an engineer would build the basic knowledge of engineering(6that would be used and the product development and
1 answer:
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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:
The load reflection coefficient,
Reflection coefficient at input,
SWR = 4.26
Given:
Characteristic impedance of the co-axial cable,
Length of the cable, L = 2.0 cm = 0.02 m
Dielectric constant, K = 2.56
frequency, f = 3.0 GHz =
Explanation:
In order to calculate the reflection coefficient at load, we first calculate these:
The line input impedance is given by:
(1)
Now, we calculate the value of :
(since, )
Now, Substituting the value in eqn (1):
Now, the load reflection coefficient is given by:
Thus
Similarly,
Reflection coefficient at input:
Now, the SWR is given by:
SWR, Standing Wave Ratio =
SWR =