The frequency at which the filter will be resonant is 1. 59 × 10^6 Hz
<h3>How to determine the frequency</h3>
The resonant frequency of the series RLC circuit is given as;
Frequency = 1/2π√(LC)
It means that the current will peak at the resonant frequency for both inductor and capacitor.
Substitute the values
Frequency = 1/2 × 3. 142 √ (50 × 10^-3 × 2 × 10^ -6)
Frequency = 1/2 × 3. 142 × 1 × 10 ^-7
Frequency = 1/ 6. 284 × 10^-7
Frequency = 1. 59 × 10^6 Hz
Thus, the frequency at which the filter will be resonant is 1. 59 × 10^6 Hz
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Answer:
A = 35h.
Area = 945 in^2.
Step-by-step explanation:
A = bh
When the base b = 35:
A = 35h.
When h = 27 ins. A = 35*27
= 945 in^2.
Answer:
Step-by-step explanation:
c sorry if im wrong
Sorry if i didn't get it right but i couldn't really tell where the two solutions seperate so....
10+y=5x+x2
y=6x-10
and
5x+y=1
y=5x+1