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When I was calculating the following netwerk:

enter image description here

Where given was: R=5,L=1,C=4,V_C(t)=10,I_T(0)=-2,V_in(t)=20.

So the ODE that we get was:

$$ \begin{cases} 20\text{V}'_{C}(t)+\text{V}_{C}(t)+4\text{V}''_{C}(t)=20\\ \text{V}_{C}(0)=10\\ 4\text{V}'_{C}(0)=-2 \end{cases} $$

Solving this ODE, gives me:

$$\text{V}_{C}(t)=20-\frac{40\cosh\left[t\sqrt{6}\right]+17\sqrt{6}\sinh\left[t\sqrt{6}\right]}{4e^{\frac{5t}{2}}}\space\space\space\text{V}$$


QUESTION:

If I plot my found solution, than on the t-axis we get a line, but in which scale? Is it seconds, minutes? or what

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1 Answer 1

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The t-axis is in seconds, assuming your argument is in seconds. If you do the unit analysis, you will find that the units you get are seconds, because seconds are the base SI unit of time.

For example:

The SI base of a farad: \$ s^4 * A^2 * m^{-2} * kg^{-1} \$
The SI base of an ohm: \$ kg * m^2 * s^{-3} * A^{-2} \$

Therefore, 1 ohm multiplied by 1 farad is 1 second.

The expression itself has implicit units, and if you wrote all the units out within the expression, you could use whatever time units you wish, you would simply have to evaluate with two different time units within the expression.

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