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You have a 192 −Ω−Ω resistor, a 0.409 −H−H inductor, a 4.95 −μF−μF capacitor, and a variable-frequency ac source with an amplitude of 3.02 VV . You connect all four elements together to form a series circuit.

(a) At what frequency will the current in the circuit be greatest? What will be the current amplitude at this frequency?
(b) What will be the current amplitude at an angular frequency of 400 rad/s? At this frequency, will the source voltage lead or lag the current?

Respuesta :

Kazeemsodikisola

Answer:

Explanation:

Given that,

The resistance of the resistor is

R = 192 Ω

The inductance of the inductor is

L = 0.409 H

The capacitance of the capacitor is

C = 4.95 μF

a. At what frequency is current max?

The current will be maximum at resonance, and resonance frequency is given as

f = 1/2π√L•C

f = 1/2π√(0.409×4.95×10^-6)

f = 111.86Hz

b. Currency at frequency 400rad/s.

w = 400rad/s

So, we need to find the inductive reactance

XL = wL

XL = 400 ×0.409

XL = 163.6 ohms

We also need the capacitive reacttance

XC = 1/wC

XC = 1/400×4.95×10^-6

XC = 505.05 ohms

Then, the impedance of the circuit is given as

Z = √(R²+(XL-XC)²)

Z = √(192²+(505.05—163.6)²)

Z = √(192²+341.9²)

Z = 392.12 ohms

Then, the current that flows can be calculated using

V= IZ

I = V/Z

I = 3.02/392.12

I = 7.7 × 10^-3 A

I = 7.7 mA

Since Xc>XL, then, the source voltage lags the current

(A). When The current will be maximum at resonance, and the resonance frequency is f = 111.86Hz

(B) When The Currency at frequency 400rad/s is I = 7.7 mA

What frequency is the current max?

Then The resistance of the resistor is

Then R is = 192 Ω

After that The inductance of the inductor is

Now, L is = 0.409 H

Then The capacitance of the capacitor is

Then C is = 4.95 μF

a. When The current will be maximum at resonance, and the resonance frequency is given as

After that f is = 1/2π√L•C

Then f is = 1/2π√(0.409×4.95×10^-6)

Now f is = 111.86Hz

b. When the Currency at frequency 400rad/s.

Then w is = 400rad/s

Then we need to find the inductive reactance

After that XL is = wL

XL is = 400 ×0.409

Then XL is = 163.6 ohms

Then We also need the capacitive reactance

XC is = 1/wC

XC is = 1/400×4.95×10^-6

XC is = 505.05 ohms

Then, the impedance of the circuit is given as

Z is = √(R²+(XL-XC)²)

Z is = √(192²+(505.05—163.6)²)

Z is = √(192²+341.9²)

Z is = 392.12 ohms

Then, the current flows can be calculated using are

V is = IZ

Then I = V/Z

I is = 3.02/392.12

I is = 7.7 × 10^-3 A

I is = 7.7 mA

Therefore, Xc>XL, then, the source voltage lags the current

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