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- 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
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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