Question

A series AC circuit contains a resistor, an inductor of 150 mH, a capacitor of 5.00 mF, and a generator with DVmax 5 240 V operating at 50.0 Hz. The maximum current in the circuit is 100 mA. Calculate (a) the inductive reactance, (b) the capacitive reactance, (c) the impedance, (d) the resistance in the circuit, and (e) the phase angle between the current and the generator voltage.

Answer 1

Given:

Inductance, L = 150 mH

Capacitance, C = 5.00 mF

`\Delta V_(max)` = 240 V

frequency, f = 50Hz

`I_(max)` = 100 mA

Solution:

To calculate the parameters of the given circuit series RLC circuit:

angular frequency,
`\omega` =
`2\pi f = 2\pi *50 = 100\pi`

a). Inductive reactance,
`X_(L)` is given by:

`\X_(L) = \omega L = 100\pi * 150* 10^(-3) = 47.12\Omega`

`X_(L) = 47.12\Omega`

b). The capacitive reactance,
`X_(C)` is given by:

`\X_(C) = (1)/(\omega C) = (1)/(2\pi fC) = (1)/(2\pi * 50* 5.00* 10^(-3)) = 0.636\Omega`

`X_(C) = 0.636\Omega`

c). Impedance, Z =
`(\Delta V_(max))/(I_(max)) = (240)/(100* 10^(-3)) = 2400\Omega`

`Z = 2400\Omega`

d). Resistance, R is given by:

`Z = \sqrt {R^(2) + (X_(L) - X_(C))}`

`2400^(2) = R^(2) + (47.12 - 0.636)^(2)`

`R = \sqrt {5757839.238}`

`R = 2399.5\Omega`

e). Phase angle between current and the generator voltage is given by:

`tan\phi = (X_(L) - X_(C))/(R)`

`\phi =tan^(-1)( (X_(L) - X_(C))/(R))`

`\phi =tan^(-1)( (47.12 - 0.636)/(2399.5)) = tan^(-1){0.0.01937}`

`\phi = 1.11^(\circ)`