Can you be a bit more specific plz and that will let me identify the answer
Answer:
la escuela,en casa y listo...............
Given:
frequency, f = 60.0 Hz
frequency, f' = 45.0 Hz
Solution:
To calculate max current in inductor, 
:
At f = 60.0 Hz
L = 0.1326 H
Now, reactance 
at f' = 45.0 Hz:
Now, is given by:
Therefore, max current in the inductor, = 2.13 A
Answer: A capacitor connected across the output allows the AC signal to pass through it and blocks the DC signal, thus acting as a high pass filter. The output across the capacitor is thus an unregulated filtered DC signal. This output can be used to drive electrical components like relays, motors, etc.
Explanation:
Answer:
8.85 Ω
Explanation:
Resistance of a wire is:
R = ρL/A
where ρ is resistivity of the material,
L is the length of the wire,
and A is the cross sectional area.
For a round wire, A = πr² = ¼πd².
For aluminum, ρ is 2.65×10⁻⁸ Ωm, or 8.69×10⁻⁸ Ωft.
Given L = 500 ft and d = 0.03 in = 0.0025 ft:
R = (8.69×10⁻⁸ Ωft) (500 ft) / (¼π (0.0025 ft)²)
R = 8.85 Ω
