Question

A series AC circuit contains a resistor, an inductor of 240 mH, a capacitor of 5.40 µF, and a source with ΔVmax = 240 V operating at 50.0 Hz. The maximum current in the circuit is 170 mA.

(a) Calculate the inductive reactance.

_________________Ω

(b) Calculate the capacitive reactance.

______________-Ω

(c) Calculate the impedance.

_______________kΩ

(d) Calculate the resistance in the circuit.

______________kΩ

(e) Calculate the phase angle between the current and the source voltage.

_______________°

Answer 1

Given that

A series AC circuit contains a inductor having inductance, L = 240 mH = 0.240 H,

a capacitor having capacitance, C = 5.40 =5.40×10^-6 F

And a resistor having resistance, R ( assumed)

Source maximum voltage, Vmax =240 V

And frequency , f = 50 Hz

The maximum circuit current,

Imax = 170 mA = .170 A

Now,

(a) we know that for AC circuit the inductive reactance is X_{​​​​​​L} = ( where, =2πf,

f= frequency and L = inductance)

Or, X_{​​​​​​L} = 2π×50×0.240 = 75.398

So, the inductive reactance is 75.398

(b) we know that for AC circuit the capacitive reactance is,    

( Where C= capacitance and w = 2πf)

or, X_{​​​​​​C}​​​ = 1/(2π×50×5.40× 10^-6)

= 589.462  

So, the capacitive reactance is 589.462  

(c) We know that for LCR AC circuit the inpedence,   

Or, Z = 240/.170 = 1411.764 = 1.4117

So, the inpedence of the circuit is 1.4117 .

(d) We know that

Or,

Or, R​​​​​^{​​​​​​2}= 1411.7² - ( 75.398 - 589.462)²

^{Or , R = 1314.775 = 1.3147}  

^{So, the resistance of resistor is 1.1314}

(e) The phase angle between current and source Voltage is ​​​​​​​