Given the circuit at the right in which the following values are used: R1 = 20 kΩ, R2 = 12 kΩ, C = 10 µ F, and ε = 25 V. You clo
se the switch at t = 0. Find (a) the current in R1 and R2 at t=0, (b) the voltage across R1 after a long time. (Careful with this one.)
1 answer:
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
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This question is not complete, the complete question is;
The stagnation chamber of a wind tunnel is connected to a high-pressure air bottle farm which is outside the laboratory building. The two are connected by a long pipe of 4-in inside diameter. If the static pressure ratio between the bottle farm and the stagnation chamber is 10, and the bottle-farm static pressure is 100 atm, how long can the pipe be without choking? Assume adiabatic, subsonic, one-dimensional flow with a friction coefficient of 0.005
Answer:
the length of the pipe is 11583 in or 965.25 ft
Explanation:
Given the data in the question;
Static pressure ratio; p1/p2 = 10
friction coefficient f = 0.005
diameter of pipe, D =4 inch
first we obtain the value from FANN0 FLOW TABLE for pressure ratio of ( p1/p2 = 10 )so
4fL / D = 57.915
we substitute
(4×0.005×L) / 4 = 57.915
0.005L = 57.915
L = 57.915 / 0.005
L = 11583 in
Therefore, the length of the pipe is 11583 in or 965.25 ft
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
Calculate the equivalent capacitance of the three series capacitors in Figure 12-1
a) 0.01 μF
b) 0.58 μF
c) 0.060 μF
d) 0.8 μF
Answer:
The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF
Therefore, the correct option is (c)
Explanation:
Please refer to the attached Figure 12-1 where three capacitors are connected in series.
We are asked to find out the equivalent capacitance of this circuit.
Recall that the equivalent capacitance in series is given by
Where C₁, C₂, and C₃ are the individual capacitance connected in series.
C₁ = 0.1 μF
C₂ = 0.22 μF
C₃ = 0.47 μF
So the equivalent capacitance is
Rounding off yields
The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF
Therefore, the correct option is (c)
Answer:
The value of v2 in each case is:
A) V2=3v for only Vs1
B) V2=2v for only Vs2
C) V2=5v for both Vs1 and Vs2
Explanation:
In the attached graphic we draw the currents in the circuit. If we consider only one of the batteries, we can consider the other shorted.
Also, what the problem asks is the value V2 in each case, where:
If we use superposition, we passivate a battery and consider the circuit affected only by the other battery.
In the first case we can use an equivalent resistance between R2 and R3:
And
In the second case we can use an equivalent resistance between R2 and (R1+R4):
And
If we consider both batteries:
