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]
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
important values.
6 kN
10 KN
2 kN/m
B
D
1 m
1 m
1 m
1 m
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
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