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Nonamiya [84]
3 years ago
11

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.)

Engineering
1 answer:
agasfer [191]3 years ago
8 0

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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Applying the Entropy Balance: Closed Systems Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston–
Mrrafil [7]

Answer:

a) the amount of energy produced in kJ/K is 0.73145 kJ/K

b) the amount of energy produced in kJ/K is 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

Explanation:

Draw the T-s diagram.

a)

C_p = 0.939 kJ/kg.K , m = 5 kg , T₂ = 520 K , T₁ = 280

R = [8.314 kJ / 44.01 kg.K] , P₂ = 20 bar , P₁ = 2 bar

Δs = m[c_p ln(\frac{T_2}{T_1}) - Rln(\frac{P_2}{P_1})]

Substitute all parameters in the equation

Δs = 5[(0.939) ln(\frac{520}{280}) - (\frac{8.314}{44.01})ln(\frac{20}{2})]

Δs = 5 kg × 0.14629 kJ/kg.K

    = 0.73145 kJ/K

b)

Δs = m[\frac{s^0(T_2) - s^0(T_1)}{M} - Rln(\frac{P_2}{P_1})]

Where T₁ = 280 K , s°(T₁) = 211.376 kJ/kmol.K

           T₂ = 520 K , s°(T₂) = 236.575 kJ/kmol.K

R = [8.314 kJ / 44.01 kg.K] , M = 44.01 kg.K , P₂ = 20 bar , P₁ = 2 bar

Δs = 5[\frac{236.575 - 211.376}{44.01} - (\frac{8.314}{44.01})ln(\frac{20}{2})]

    = 5 kg (0.13795 kJ/kg.K)

    = 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

7 0
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Consider the diffusion of water vapor through a polypropylene (PP) sheet 1 mm thick. The pressures of H2O at the two faces are 3
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Answer:

\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}

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Therefore, the diffusion flux is \boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}

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