Answer:
il(t) = e^(-100t)
Explanation:
The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.
The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...
il(t) = e^(-t/.01)
il(t) = e^(-100t) . . . amperes
Answer:
Explanation:
ADT for an 2-D array:
struct array{
int arr[10];
}arrmain[10];
An application that stores an array with 1000 rows and 1000 columns, where less than 10,000 of the array values are non-zero. The two different implementations for such arrays that would be more space efficient than a standard two-dimensional array implementation requiring one million positions are :
1) struct array{
int *p;
}arr[1000];
2) struct array{
int *p;
}arr[1000];
Answer: 255
255 turns are required to create 25 ohms of secondary impedance.
Explanation:
Given that,
Number of turns in primary wire N₁ = 900
impedance on Primary wire Z₁ = 400 ohms
Number of turns in Secondary wire N₂ = ?
impedance on Secondary wire Z₂ = 25 ohms
we know that, the relationship between turn and impedance is
Zp / Zs = ( Np / Ns )²
(Primary impedance / secondary impedance) = Number of turns in primary wire / Number of turns in secondary wire)²
there fore
Z₁ / Z₂ = ( N₁ / N₂ )²
Now we substitute
( 400 / 25 ) = ( 900 / N₂ )²
400 / 25 = 900² / N₂²
we cross multiple to get our N₂
400 × N₂² = 900² × 25
N₂² = ( 900² × 25 ) / 400
N₂² = ( 810000 × 25 ) / 400
N₂² = 20250000 / 400
N₂² = 50625
N₂ = √50625
N₂ = 225
Therefore 255 turns are required to create 25 ohms of secondary impedance.