Answer:
a) the required time is 0.6283 μs
b) the inductor current is 0.5 mA
Explanation:
Given the data in the question;
The capacitor voltage has its maximum value of 25 V at t = 0
i.e V[tex]_m[/tex] = V₀ = 25 V
we determine the angular velocity;
ω = 1 / √( LC )
ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )
ω = 1 / √( 1.6 × 10⁻¹³ )
ω = 1 / 0.0000004
ω = 2.5 × 10⁶ s⁻¹
a) How much time does it take until the capacitor is fully discharged for the first time?
V[tex]_m[/tex] = V₀sin( ωt )
we substitute
25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )
25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )
divide both sides by 25 V
sin( 2.5 × 10⁶ × t ) = 1
( 2.5 × 10⁶ × t ) = π/2
t = 1.570796 / (2.5 × 10⁶)
t = 0.6283 × 10⁻⁶ s
t = 0.6283 μs
Therefore, the required time is 0.6283 μs
b) What is the inductor current at that time?
[tex]I[/tex](t) = V₀√(C/L) sin(ωt)
{ sin(ωt) = 1 )
[tex]I[/tex](t) = V₀√(C/L)
we substitute
[tex]I[/tex](t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )
[tex]I[/tex](t) = 25 × 0.00002
[tex]I[/tex](t) = 0.0005 A
[tex]I[/tex](t) = 0.5 mA
Therefore, the inductor current is 0.5 mA
The time taken for the capacitor to fully discharge is 6.28 x 10⁻⁷ s.
The current in the inductor at the given time is 0.0005 A.
The angular velocity of the circuit is calculated as follows;
[tex]\omega = \frac{1}{\sqrt{LC} } \\\\\omega = \frac{1}{\sqrt{20 \times 10^{-3} \times 8 \times 10^{-12} } } \\\\\omega = 2.5 \times 10^6 \ rad/s[/tex]
V = V₀(sinωt)
sinωt = V/V₀
sinωt = = 25/25
sin(ωt) = 1
ωt = sin⁻¹ (1)
ωt = π/2
2.5 x 10⁶ t = π/2
t = 1.57 / (2.5 × 10⁶)
t = 6.28 x 10⁻⁷ s
The current in the inductor at the given time is calculated as follows
[tex]I(t) = V_0 \sqrt{\frac{C}{L} } \\\\I = 25 \times \sqrt{\frac{8\times 10^{-12}}{20 \times 10^{-3}} } \\\\I = 0.0005 \ A[/tex]
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