Two technicians are discussing electromagnetic induction. Technician A says that the induced voltage can be increased if the spe
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initial speed of 226000 m/s
acceleration of 4.0 x 1014 m/s2,
speed of 781000 m/s
What is Acceleration?
- Acceleration is a rate of change of velocity with respect to time with respect to direction and speed.
- A point or an object moving in a straight line is accelerated if it speeds up or slows down.
- Acceleration formula can be written as,
a = (v - u ) / t m/s²
As we have to find the time taken, the formula can be altered as,
where, t - time taken to reach a final speed
v - final velocity
u - initial velocity
a - acceleration.
Substituting all the given values,
= 1.3875 × 10⁻⁹ seconds.
So, taken to reach the final speed is found to be 1.3 × 10⁻⁹ 8iH..
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 = 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 = 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?
(t) = V₀√(C/L) sin(ωt)
{ sin(ωt) = 1 )
(t) = V₀√(C/L)
we substitute
(t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )
(t) = 25 × 0.00002
(t) = 0.0005 A
(t) = 0.5 mA
Therefore, the inductor current is 0.5 mA