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If you already know the explicit form of the function $f(x)$, then it's possible to find the periode by solving this equation $f(x) = f(x+P)$. If you have a graphical representation of the function (finite), maybe by collecting the numerical values or other means then you have to follow the next two steps: Step one: Trivial extension Consider the ...

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If you in the USA there are couple of good ways to start building a profile. Volunteer at your local FIRST robotics team. This is good place to get started learning new skills and using current skills Be an active member of the communities such as engineering stackexchange where you can start building an online profile of you skills. You can asked embedded ...

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Because it is not a power cable (Q1) using power connectors (Q2). BNC cable has a solid conductor and a ground braid to electrically shield and mechanically protect the conductor. Electrical reason: Some ground braids are made of steel, which is not as good a conductor as copper, so voltage losses to the wire would be higher than copper wire. Basically, ...

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Your circuit works. Start at your -8V and 0mA. You know the voltage. $V_{R_1} = 0 - (-8V) = 8V$. All other load currents must flow through it (to get back to battery). $I_{R_1} = 6mA + 18mA + 4mA = 28mA = I_T$. Calculate $R_1$. For 4V at 4mA. Voltage: $V_{R_2}$ = 4V - 0V = 4V. (This may seem obvious, but you need to do this for the following ...

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$R_4 \parallel R_5$ and $R_3$ form a voltage divider: $$V_A = \frac {R_4 \parallel R_5}{R_4 \parallel R_5 + R_3} \times (V_1 - V_2)$$ Or alternatively using series/parallel rules and Ohm's Law: $$I_A = \frac {V_1 - V_2}{R_4 \parallel R_5 + R_3}$$ $$V_A = I_A \ (R_4 \parallel R_5)$$

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An induction heater creates an oscillating magnetic field near the coil. The "mechanism" to damage an electronic component is that eddy currents are induced in the metal parts of the component. These eddy currents heat up the part until something breaks. If the part breaks, yes, it will be due to magnetic induction.

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There is not enough space. But they do manage to find enough space on the bigger ceramic components though.

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If you consider the RHS loop of your circuit and using $i_3$ as the current going through $R_3$ (in the same direction as $i$), $i_4$ the current going through $R_4$ and $i_5$ the current going through $R_5$, you have (assuming your polarity of $V_1$ and $V_2$ is correct):  -V_2 + V_1 + R_3 i_3 - V_A = 0\\ V_A = -R_4 i_4\\ V_A = -R_5 i_5\\ i_3 = i_4 + ...

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$R_{23} = {\frac{1}{\frac{1}{R_2}+\frac{1}{R_3}}}$ Resistors in parallel $R_{23} \approx 9.018KΩ$ Plugging in the values $V_{out} = V_{in} \cdot {\frac {R_{23}}{R_{1} + R_{23}}}$ Voltage division $V_{out} \approx 0.9895 \cdot V_{in}$ Plugging in the values $k \approx 0.9895$

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Assume your design works, so you know that the voltages at A, B, and so on are the ones given in the problem. Since you know the currents through the loads, you know some of the currents flowing out of the nodes A, B, C .... , With knowledge of the voltages, and some of the currents, you can figure out the other currents. Then you can solve for the ...

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It would not be a good idea at all - a transformer is effectively a magnetic circuit, so you need a low resistance path for the magnetic flux to flow through. Copper has a high magnetic reluctance, so would not be a good choice. https://en.wikipedia.org/wiki/Magnetic_reluctance https://en.wikipedia.org/wiki/Permeability_(electromagnetism)#...

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