Question

If an R = 1-kΩ resistor, a C = 1-μF capacitor, and an L = 0.2-H inductor are connected in series with a V = 150 sin (377t) volts source, what is the maximum current delivered by the source?

Answer 1

Answer

given,

R = 1-kΩ  = 1000 Ω

C = 1-μF

L = 0.2-H

V = V_max sin( ω t)

comparing

V = 150 sin ( 377 t)

ω = 377

$\chi_c = \dfrac{1}{\omega C}$

$\chi_c = \dfrac{1}{377 \times 1 \times 10^{-6}}$

$\chi_c = 2652.5\Omega$

$\chi_L =377 \times 0.2$

$\chi_L =75.4\ \Omega$

Impedance,

$Z = \sqrt{R^2+(\chi_L-\chi_c)^2}$

$Z = \sqrt{1000^2+(75.4 -2652.5)^2}$

Z = 2764.3 Ω

now,

V_{max} = 150 V

$I_{max} = \dfrac{V}{Z}$

$I_{max} = \dfrac{150}{2764.3}$

$I_{max} = 0.0543$

$I_{max} = 54.3\ mA$