Answer:
Explanation:
Inductance L = 1.4 x 10⁻³ H
Capacitance C = 1 x 10⁻⁶ F
a )
current I = 14 .0 t
dI / dt = 14
voltage across inductor
= L dI / dt
= 1.4 x 10⁻³ x 14
= 19.6 x 10⁻³ V
= 19.6 mV
It does not depend upon time because it is constant at 19.6 mV.
b )
Voltage across capacitor
V = ∫ dq / C
= 1 / C ∫ I dt
= 1 / C ∫ 14 t dt
1 / C x 14 t² / 2
= 7 t² / C
= 7 t² / 1 x 10⁻⁶
c ) Let after time t energy stored in capacitor becomes equal the energy stored in capacitance
energy stored in inductor
= 1/2 L I²
energy stored in capacitor
= 1/2 CV²
After time t
1/2 L I² = 1/2 CV²
L I² = CV²
L x ( 14 t )² = C x ( 7 t² / C )²
L x 196 t² = 49 t⁴ / C
t² = CL x 196 / 49
t = 74.8 μ s
After 74.8 μ s energy stored in capacitor exceeds that of inductor.
Answer:
The fence is 5feet less.
Explanation:
We need to determine
The less amount of fence required, if the enclosure has full width and reduced length, compared to full length and reduced width.
Approach & WorkingArea of lawn = 30 × 403/4th of the area of lawn = ¾(30 × 40) = 30 * 30
When full width will be fenced, and reduced length will be fenced.
Width = 30 feet30 * L = 30 * 30Hence, length = 30 feetLength of fence needed = 2(30 + 30) = 120 feet
When full length will be fenced, and reduced width will be fenced
Length = 40 feet40 * W = 30 * 30W = 22.5 feetLength of fence needed = 2(40 + 22.5) = 125 feet
Difference in length of fence needed = 125 – 120 = 5 feet.
Answer:
C
Explanation:
Because this same question was on my test last week and I got it correct
