Question

A 6.00-mH solenoid is connected in series with a 5.0-μF capacitor and an AC source. The solenoid has internal resistance 3.0 Ω which can be treated as a series resistor in the circuit.Part A: What is the resonant angular frequency?Part B: If the source emf amplitude is Emax = 69 V what is the current amplitude at the resonant angular frequency?

Answer 1

Answer:

5773.50269 Hz

23 A

Explanation:

$L$ = Inductance = 6 mH

$C$ = Capacitance = 5 μF

$R$ = Resistance = 3 Ω

$\epsilon$ = Maximum emf = 69 V

Resonant angular frequency is given by

$\omega=\dfrac{1}{\sqrt{LC}}\\\Rightarrow \omega=\dfrac{1}{\sqrt{6\times 10^{-3}\times 5\times 10^{-6}}}\\\Rightarrow \omega=5773.50269\ Hz$

The resonant angular frequency is 5773.50269 Hz

Current is given by

$I=\dfrac{\epsilon}{R}\\\Rightarrow I=\dfrac{69}{3}\\\Rightarrow I=23\ A$

The current amplitude at the resonant angular frequency is 23 A

Answer 2

a. The resonant angular frequency is 5773.50269 Hz

b. The current amplitude at the resonant angular frequency is 23A.

Calculation of the frequency & current amplitude:

Since

L =  Inductance = 6 mH

C = Capacitance = 5 μF

R = Resistance = 3 Ω

e = Maximum emf = 69 V

a.

So here the angular frequency is

= 1 / √LC

= 1 / √6*10^-3*5*10^-6

= 5773.50269 Hz

b. Now th current amplitude is

= e/R

= 69/3

= 23A

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