Answer:
- The resistance of the circuit is 1250 ohms
- The 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 (
) is equal to inductive reactance (
),
Where;
R is the resistance of the circuit, calculated as;
![R = \frac{V}{I} \\\\R = \frac{250 \ \times \ 10^{-3}}{0.2 \ \times \ 10^{-3}} \\\\R = 1250 \ ohms](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7BV%7D%7BI%7D%20%5C%5C%5C%5CR%20%3D%20%5Cfrac%7B250%20%5C%20%5Ctimes%20%5C%2010%5E%7B-3%7D%7D%7B0.2%20%5C%20%5Ctimes%20%5C%2010%5E%7B-3%7D%7D%20%5C%5C%5C%5CR%20%3D%201250%20%5C%20ohms)
The inductive reactance is calculated as;
![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\\](https://tex.z-dn.net/?f=X_l%20%3D%20X_c%20%3D%20%5Cfrac%7B1%7D%7B%5Comega%20C%7D%20%3D%20%5Cfrac%7B1%7D%7B2%5Cpi%20f_o%20C%7D%20%3D%20%5Cfrac%7B1%7D%7B2%5Cpi%20%28100%5Ctimes%2010%5E3%29%280.04%5Ctimes%2010%5E%7B-6%7D%20%29%20%7D%20%3D%2039.789%20%5C%20ohms%5C%5C)
The inductance is calculated as;
![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](https://tex.z-dn.net/?f=X_l%20%3D%20%5Comega%20L%20%3D%202%5Cpi%20f_o%20L%5C%5C%5C%5CL%20%3D%20%5Cfrac%7BX_l%7D%7B2%5Cpi%20f_o%7D%5C%5C%5C%5CL%20%3D%20%5Cfrac%7B39.789%7D%7B2%5Cpi%20%28100%20%5Ctimes%2010%5E3%29%7D%20%20%5C%5C%5C%5CL%3D%206.3%20%5C%20%5Ctimes%20%5C%2010%5E%7B-5%7D%20%5C%20H%5C%5C%5C%5CL%20%3D%200.063%20%5Ctimes%20%5C%2010%5E%7B-3%7D%20%5C%20H%5C%5C%5C%5CL%20%3D%200.063%20%5C%20mH)
Answer:
The answer is below
Explanation:
Let A represent the first switch, B represent the second switch and C represent the bulb. Also, let 0 mean turned off and 1 mean turned on. Since when both switches are in the same position, the light is off. This can be represented by the following truth table:
A B C (output)
0 0 0
0 1 1
1 0 1
1 1 0
The logic circuit can be represented by:
C = A'B + AB'
The output (bulb) is on if the switches are at different positions; if the switches are at the same position, the output (bulb) is off. This is an XOR gate. The gate is represented in the diagram attached below.
Answer:
The field strength needed is 0.625 T
Explanation:
Given;
angular frequency, ω = 400 rpm = (2π /60) x (400) = 41.893 rad/s
area of the rectangular coil, A = L x B = 0.0611 x 0.05 = 0.003055 m²
number of tuns of the coil, N = 300 turns
peak emf = 24 V
The peak emf is given by;
emf₀ = NABω
B = (emf₀ ) / (NA ω)
B = (24) / (300 x 0.003055 x 41.893)
B = 0.625 T
Therefore, the field strength needed is 0.625 T
I think it’s manufacturing
You need to explain it more simple as everyone is clueless