Question

An L-R-C series circuit consists of a 60.0 Ω resistor, a 10.0 μF capacitor, a 3.60 mH inductor, and an ac voltage source of voltage amplitude 60.0 V operating at 1450 Hz .

a.) Find the current amplitude across the inductor, the resistor, and the capacitor.

b.) Find the voltage amplitudes across the inductor, the resistor, and the capacitor.

c.) Why can the voltage amplitudes add up to more than 60.0 V ?

d.) If the frequency is now doubled, but nothing else is changed, which of the quantities in part A and B will change?

- only current amplitude will change

-only voltage amplitude across the inductor and capacitor will change

-only voltage amplitude across the inductor will change

-current amplitude and voltage across the any circuit element will change

e.) Find new current amplitude across the inductor, the resistor, and the capacitor.

f.) Find new voltage amplitudes across the inductor, the resistor, and the capacitor.

Answer 1

a)
Inductive reactance, XL = 2*pi*f*L = 2*pi*1450*3.6*10^-3 = 32.8 ohms

Capacitive reactance, XC = 1/(2*pi*f*c) = 1/(2*pi*1450*10*10^-6) = 10.98 ohms

R = 60 ohms

Impedance , z = sqrt(R^2 + (XL - Xc)^2)

= sqrt(60^2 + (32.8 - 10.98)^2 )

= 63.84 ohms

current ampiltude in te ckt, Imax = Vmax/z

= 60/63.84

= 0.94 A

b) vR = Imax*R

= 0.94*60

= 56.4 volts

VL = XL*Imax

= 32.8*0.94

= 30.8 volts

VC = XC*Imax

= 10.98*0.94

= 10.3 volts

c) because voltage across inductor and capcitor not in phase.

d) -current amplitude and voltage across the any circuit element will change

e) when frequency is doubled,

Inductive reactance, XL = 2*pi*f*L = 2*pi*2*1450*3.6*10^-3 = 65.6 ohms

Capacitive reactance, XC = 1/(2*pi*f*c) = 1/(2*pi*2*1450*10*10^-6) = 5.49 ohms

R = 60 ohms

Impedance , z = sqrt(R^2 + (XL - Xc)^2)

= sqrt(60^2 + (65.6 - 5.49)^2 )

= 84.93 ohms

current ampiltude in te ckt, Imax = Vmax/z

= 60/84.93

= 0.706 A

f) vR = Imax*R

= 0.706*60

= 42.36 volts

VL = XL*Imax

= 65.6*0.706

= 46.3 volts

VC = XC*Imax

= 5.49*0.706

= 3.88 volts