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In my Circuit Theory lesson, we were studying RC circuits and there was a question which I tried to make before our senior but I accidentally came up with a different answer due to choosing current direction opposite to our senior;

schematic

simulate this circuit – Schematic created using CircuitLab

It's the basic RC structure, the switch was closed for enough time and opens at \$t=0\$;

schematic

simulate this circuit

Notice the red arrow at the right, because of its direction my calculations went this way: $$KCL\; at \; Node\; 1:$$ $$-i_R+i_C=0$$ $$-\frac{V_C}{6}+10\mu\frac{dV_C}{dt}=0$$ $$\frac{dV_C}{dt}+(-\frac{1}{60\mu}V_C)=0$$

If we solve the diferential equation according to \$\frac{dx(t)}{dt}+\alpha x(t)=0\ \Rightarrow\ x(t)=x_0e^{-\alpha t}\$ :

$$V_C(t)=V_C(0)e^{-(-\frac{1}{60\mu}t)}$$

Since \$V_C(0)=9\$ :

$$V_C(t)=9e^{\frac{1}{60\mu}t}$$

And here it is, according to this result, the voltage at C1 is increasing over time which is impossible. My senior's result was \$V_C(t)=9e^{-\frac{1}{60\mu}t}\$ because he took the direction of the \$i_C\$ opposite to mine, therefore at KCL, signs of currents were the same so the final equation had a negative sign, unlike my result.

I've been told that I can choose the current and polarization directions as I want at Node or Mesh analysis and it will not affect the final results(maybe voltage and current signs can be wrong) but in this case, it changed the whole equation to something meaningless.

I would like to know what I am missing here.

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1 Answer 1

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The basic constitutive equation for a capacitor is based on these assumed polarities,

enter image description here

The equation being, $$i(t)=C\frac{dv}{dt}$$ If you want to define your current as leaving the capacitor then you need to include a negative sign in your dv/dt term.

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    \$\begingroup\$ Yes, good catch. So a simple way to double-check this is that current entering into capacitor increases the voltage (dv positive), and current exiting the capacitor decreases the voltage (dv negative). So no matter how the current direction is selected, we know we are discharging the capacitor so dv is negative anyway. \$\endgroup\$
    – Justme
    Oct 31, 2020 at 22:56
  • \$\begingroup\$ Aha, so it is why the Passive Sign Convention is defined before these assumptions, I've been wondering why we needed that and so my answer is my question. :) \$\endgroup\$
    – Berk
    Nov 1, 2020 at 4:57

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