The total circuit current at the resonant frequency is 0.61 amps
What is a LC Circuit?
- A capacitor and an inductor, denoted by the letters "C" and "L," respectively, make up an LC circuit, also referred to as a tank circuit, a tuned circuit, or a resonant circuit.
- These circuits are used to create signals at particular frequencies or to receive signals from more complicated signals at particular frequencies.
Q =15 = (wL)/R
wL = 30 ohms = Xl
R = 2 ohms
Zs = R + jXl = 2 +j30 ohms where Zs is the series LR impedance
| Zs | = 30.07 <86.2° ohms
Xc = 1/(wC) = 30 ohms
The impedance of the LC circuit is found from:
Zp = (Zs)(-jXc)/( Zs -jXc)
Zp = (2+j30)(-j30)/(2 + j30-j30) = (900 -j60)2 = 450 -j30 = 451 < -3.81°
I capacitor = 277/-j30 = j9.23 amps
I Zs = 277/(2 +j30) = (554 - j8,310)/904 = 0.61 - j9.19 amps
I net = I cap + I Zs = 0.61 + j0.04 amps = 0.61 < 3.75° amps
Hence, the total circuit current at the resonant frequency is 0.61 amps
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I think the answer is ruthorford
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<h2>
Answer:</h2>
The rate of deceleration is -0.14
<h2>
Explanation:</h2>
Using one of the equations of motion;
v = u + at
where;
v = final velocity of the boat = 0m/s (since the boat decelerates to a stop)
u = initial velocity of the boat = 25m/s
a = acceleration of the boat
t = time taken for the boat to accelerate/decelerate from u to v = 3 minutes
<em>Convert the time t = 3 minutes to seconds;</em>
=> 3 minutes = 3 x 60 seconds = 180seconds.
<em>Substitute the values of v, u, t into the equation above. We have;</em>
v = u + at
=> 0 = 25 + a(180)
=> 0 = 25 + 180a
<em>Make a the subject of the formula;</em>
=> 180a = 0 - 25
=> 180a = -25
=> a = -25/180
=> a = -0.14
The negative value of a shows that the boat is decelerating.
Therefore, the rate of deceleration of the speed boat is 0.14