RLC Circuit Analysis -- Two sources and two switches

AI Thread Summary
The discussion focuses on analyzing an RLC circuit with two sources and switches, confirming the correctness of the approach and calculations. The capacitor behaves as an open circuit for t < 0 and t = ∞, with voltage values of 9V and 5V respectively. The time constant is calculated as T = 0.2 seconds, leading to the voltage equation V(t) = 5 + 4e^{-5t}. Feedback indicates that the approach and reasoning are sound, though there is a sign error in some exponents. Overall, the analysis demonstrates a solid understanding of the series RC differential equation.
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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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Your approach and results look good.
 
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.
 
Thank You Very Much!
 
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