Maths in electircal engineering (1 Viewer)

[jm1234567890]

I'm doing a double with maths, but most people don't and do a tiny amount of maths.

As a result when they tend to just use the premade formulas to solve them. While I derive the circuit from first principles.

I think just subing into equations does not allow you to understand the circuit fully. What do you people think?

eg. try doing this problem with "formulas"

 

[wogboy]

I think just subing into equations does not allow you to understand the circuit fully. What do you people think?

I agree that it is essential for an engineer to understand how to analyse a circuit from the fundamental principles (Ohm's law, Kirchoff's laws etc) rather than only memorising & applying formulas/solutions, but you'll find that in practice if you're designing a circuit (or trying to understand an existing one), you don't want to spend a long time deriving the output from fundamentals (time=money). In that case, you should try to apply memorised results/formulas if you can (e.g. the response of RC circuits under DC excitation), so long as you completely understand the origins of those forumlas and how to apply them.

BTW, that circuit can be analysed by using just "formulas" (along with a bit of ingenuity). The voltage across that dependent source is 0.5v, so by KVL the voltage across the 8 ohm resistor is 0.5v too. So that dependent source can be transformed into an 8 ohm resistor. So the time constant of that circuit ends up as T = RC = 0.615s (combine the series & parallel resistors), so the output is v(t) = 20*e^-t/0.615 for t>=0. (no need for solving any messy algebraic or differential equations)

 

[asdf]

eg. try doing this problem with "formulas"

I'm not sure which formulas your talking about? Heres how i would do it though:

dv/dt + 1.625v = 0

from here its obvious that:
v(t) = A e^(-1.625)t <<<<----Is this the memorised formula your talking about?
if v(0) = 20 then A = 20.
v(t) = 20e^(-1.625t)

But wogboys method is obviously cooler.

 

[jm1234567890]

I'm not sure which formulas your talking about? Heres how i would do it though:

dv/dt + 1.625v = 0

from here its obvious that:
v(t) = A e^(-1.625)t <<<<----Is this the memorised formula your talking about?
if v(0) = 20 then A = 20.
v(t) = 20e^(-1.625t)

But wogboys method is obviously cooler.

"dv/dt + 1.625v = 0"
how did you get that line


but this is just a simple first order circuit. It is not as easy to plug-in solutions of DE's in 2nd order.

wogboy's method was the suggested solution by the lecturer.

 

[mlando]

as an electrical engineering/maths double person i have found that maths makes the electrical engineering much easiers. especially in later years when you are looking at signals processing, control, communications, etc...

having said that, i havent even considered designing an electric circuit without a computer since first year. understanding how to do it is great, but spending three weeks to build a simple circuit that takes two hours to design and test with SPICE is a waste of time.