A resistor ( R = 9.00 10 2 ), a capacitor ( C = 0.250 F), and an inductor ( L =

Question

A resistor

(R = 9.00 102 ),

a capacitor

(C = 0.250 F),

and an inductor

(L = 1.30 H)

are connected in series across a 2.40 102-Hz AC source for which

Vmax = 1.10 102 V.

(a) Calculate the impedance of the circuit.
k

(b) Calculate the maximum current delivered by the source.
A

(c) Calculate the phase angle between the current and voltage.
°

(d) Is the current leading or lagging behind the voltage?

The current leads the voltage.The current lags behind the voltage.     There is no phase difference between the current and voltage.

Explanation / Answer

Given,
R = 9* 10^2   = 900
C = 250 * 10^-6 F
L = 1.3H
f = 2.4 * 10^2 HZ = 240HZ

a)
Impedance (Z) = sqrt(R^2 + (XL-XC)^2)
XC = 1/2* pi* f*C
XC = 1/2* 3.14* 240 *.250 * 10^-6 = 2653.92

XL = 2* pi*f*L
XL = 2*3.14 * 240 * 1.3
XL = 1959

Substituting Values --
Z = sqrt(900^2 + (1959.36-2653.92)^2)
Z= 1136.84
Impedance of the circuit = Z= 1136.84

b)
Vmax = Imax * Z
Imax = Vmax/Z
Imax = 1.1*10^2 /1136.84
Imax = 0.0967 Amp
Maximum current delivered by the source Imax = 0.0967 Amp

c)
Let Phase angle be
cos = R/Z
cos = 900/1136.84
= COS^-1(900/1136.84)
= 37.3660
Phase angle between the current and voltage = 37.3660

d)
In our case capacitive reactance is greater than the inductive reactance, XC > XL therefore the overall circuit reactance is capacitive giving a leading phase angle. Therefore current leads the voltage.

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