The current at resonance in a series L-C-R circuit is 0.2mA. If the applied voltage is 250mV at a frequency of 100 kHz and the circuit capacitance is 0.04 microfrad . Find the circuit resistance and inductance

Answers

Answer 1

Answer:

The resistance of the circuit is 1250 ohmsThe inductance of the circuit is 0.063 mH.

Explanation:

Given;

current at resonance, I = 0.2 mA

applied voltage, V = 250 mV

resonance frequency, f₀ = 100 kHz

capacitance of the circuit, C = 0.04 μF

At resonance, capacitive reactance ([tex]X_c[/tex]) is equal to inductive reactance ([tex]X_l[/tex]),

[tex]Z = \sqrt{R^2 + (X_ l - X_c)^2} \\\\But \ X_l= X_c\\\\Z = R[/tex]

Where;

R is the resistance of the circuit, calculated as;

[tex]R = \frac{V}{I} \\\\R = \frac{250 \ \times \ 10^{-3}}{0.2 \ \times \ 10^{-3}} \\\\R = 1250 \ ohms[/tex]

The inductive reactance is calculated as;

[tex]X_l = X_c = \frac{1}{\omega C} = \frac{1}{2\pi f_o C} = \frac{1}{2\pi (100\times 10^3)(0.04\times 10^{-6} ) } = 39.789 \ ohms\\[/tex]

The inductance is calculated as;

[tex]X_l = \omega L = 2\pi f_o L\\\\L = \frac{X_l}{2\pi f_o}\\\\L = \frac{39.789}{2\pi (100 \times 10^3)} \\\\L= 6.3 \ \times \ 10^{-5} \ H\\\\L = 0.063 \times \ 10^{-3} \ H\\\\L = 0.063 \ mH[/tex]


Related Questions

For the beam loaded as shown in Fig Q2a. perform the following task:
Calculate the support reactions at A and E;
(i)
Draw the shear force diagram for the beam showing all important
values;
>
(iii)
Draw the bending moment diagram for the beam showing all
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10 KN
2 kN/m
B
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Answers

Answer:

a

Explanation:

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A horizontal poly crystalline solar panel module has to be investigated by natural cooling. For crystal silicon, the thermal coefficient approximately 0.0045/K is used. Investigate the effect of air velocity on the cooling performance of PV panels at 0-5 m/s air velocities, 25-40 ºC ambient temperatures, and 400-1000 W/ m2 solar radiation

Answers

Solution :

It is given that :

Thermal coefficient = 0.0045/K

Ambient temperature, [tex]$T_a = 25 - 40^\circ$[/tex]

air velocity, v = 0-5 m/s

Solar radiation, [tex]$G= 400-100 \ W/m^2$[/tex]

[tex]$P=50 \ W$[/tex]

Model calculations :

Cell temperature ([tex]$T_c$[/tex])

[tex]$T_c = T_a + \left(\frac{0.25}{5.7+3.8 \ v_w}\right) G$[/tex]

where [tex]$ v_w - v_a = $[/tex] wind speed / air speed

∴ [tex]$T_c = 2 \pi + \left(\frac{0.25}{5.7+3.8 \times 1}\right) \times 400$[/tex]

   [tex]$T_c = 35.526 ^\circ$[/tex]

[tex]$\Delta T = T_c -25$[/tex]

      = 35.526 - 25

      = 10.526 K

Thermal coefficient = 0.0045 x 10.526

                                = 0.04737

Pv power = [tex]$(1 -C_T) \times P \times \frac{G}{1000}$[/tex]

                [tex]$=(1 -0.04737) \times 50 \times \frac{400}{1000}$[/tex]

                = 17.0526 W

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