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:
-62
Step-by-step explanation:
y³ – 2y (y – 5) + 13
when y = -3
y³ – 2y (y – 5) + 13
=(-3)³ – 2(-3) [ (-3) – 5] + 13 (by PEDMAS, evaluate parentheses first)
=(-27) + (6) (-8) + 13 (next do multiplication)
=(-27) + (-48) + 13
= -27 - 48 + 13
= -62
It would be 10^13 which would equal 10,000,000,000,000
Step-by-step explanation:
-2x+3=-4
-2x=-7
x=7/2=3.5..
HOPE IT HELPS.