Direction of current in RC parallel circut

[jaus tail]

In the figure below in e1 1.64 why is i(t) negative. When switch is put to S1 position, the capacitor will start discharging. So current will flow from top plate of capacitor to resistor to bottom plate of capacitor.
Where did negative sign come from in eqn. 1.64?

When charging the current through capacitor is C dv/dt, flowing from top to bottom in capacitor leg. Could this have something to do with it?

 

[Orodruin]

The equation tells you that when $i$ (as defined in the figure) is positive the voltage across the conductor decreases (which also is equivalent to the charge decreasing). Why do you think there is a problem with that?

 

[jaus tail]

The equation tells you that when $i$ (as defined in the figure) is positive the voltage across the conductor decreases (which also is equivalent to the charge decreasing).

So if i is positive and the voltage across the conductor is decreasing why is minus sign in equation.
Is it like since voltage across capacitor is decreasing, dv/dt is negative? So to balance the negative dv/dt, we put negative sign in equation to make current positive? I'm confused.

 

[NascentOxygen]

It is easy to get yourself confused here. It is best to start with the standard equation that you have memorised for current into a capacitor, $i(t) = \displaystyle{C\cdot\frac {dv} {dt}}$[/color] and then take its negative because we need the equation for current in the opposite direction.

 

[jaus tail]

Thanks. That was clear...

 

[Orodruin]

It is easy to get yourself confused here. It is best to start with the standard equation that you have memorised for current into a capacitor, $i(t) = \displaystyle{C\cdot\frac {dv} {dt}}$ and then take its negative because we need the equation for current in the opposite direction.

I would say even this is not enough. You also need to define what you consider to be positive charge on the capacitor. This will also define what you should consider to be a positive voltage and make it much easier to understand what direction of current increases the charge and which decreases it.

 

[NascentOxygen]

I would say even this is not enough. You also need to define what you consider to be positive charge on the capacitor. This will also define what you should consider to be a positive voltage and make it much easier to understand what direction of current increases the charge and which decreases it.

You do need to consider voltage, certainly. But the voltage polarity (or, rather, what we denote as the voltage polarity) is already defined once the direction of current is defined; the two are related by the standard D.E. They are not independent: positive current in contributes positive voltage increment.

 

[Orodruin]

They are not independent: positive current in contributes a positive voltage increment.

This is not so well defined. You still have to define what "in" means. What comes in on one side goes out on the other.

 

[NascentOxygen]

This is not so well defined. You still have to define what "in" means. What comes in on one side goes out on the other.

The direction of the current's arrow indicates "in". Once you've drawn one arrow, whether it be i_C(t) or v_C(t), everything else follows along.

 

[Orodruin]

The direction of the current's arrow indicates "in". Once you've drawn one arrow, whether it be i_C(t) or v_C(t), everything else follows along.

In that case you would get the wrong sign in this problem, because the voltage v across the capacitor is defined in the opposite way (which is why there is a minus sign).

In general, I would advocate against using "standard" formulas and for thinking.

 

[NascentOxygen]

Capacitor voltage and current are related by a standard D.E., in the same way as resistor voltage and current are related by one Ohm's Law. If those polarities don't suit the notation of the schematic then a negative sign can readily be assigned as appropriate.

Everyone should adopt techniques which succeed best for them.