RLC Circuit Analysis -- Two sources and two switches

  • #1
73
10
Homework Statement:
Given the following circuit with the source voltages V1=9(V) and V2=5(V). Switch 1 has been connected to A and switch 2 has been closed for a long time. At t=0, switch 1 is connected to B, and switch 2 is open. Find the constants α, β, and γ in the expression of the voltage v(t) through the capacitor

V(t) = α + βe^(γt) V
Relevant Equations:
$$V(t) = V( ∞) + [V(0) - V( ∞)] e ^ {\frac{t}{T}}$$
Hello, this is my working. My professor did not give any answer key, and thus can I check if I approach the question correctly, and also check if my answer is correct at the same time.

When t < 0, capacitor acts as open circuit,
$$V(0-) = V(0+) = 9V$$

When t = infinity,
$$V( ∞) = 5V$$ (because switch is now connected to B, and capacitor acts as open circuit when t = ∞)

Time Constant, $$T = RC = 2 * 100 * 10^{-3} = 0.2 s$$

When t > 0,
$$V(t) = V( ∞) + [V(0) - V( ∞)] e ^ {\frac{t}{T}}$$
$$V(t) = 5 + [9 - 5] e^{\frac{-t}{0.2}}$$
$$ V(t) = 5 + 4e^{-5t}$$

Therefore,
α = 5,
β = 4
γ = -5
 

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Answers and Replies

  • #2
gneill
Mentor
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Your approach and results look good.
 
  • #3
DaveE
Science Advisor
Gold Member
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Looks good. You have a sign error in some of your exponents.
Your reasoning is great; the simple approach. There are much harder ways to solve this that you successfully avoided. Of course it does rely on you already knowing the solution to the series RC differential equation (i.e. time constant and exponentials), but that seems reasonable since they gave you the form of the answer.
 
  • #4
73
10
Thank You Very Much!
 

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