MultiMeter Help!

42.

Duh.

 

Well actually it's 41.99978969878979, but close enough I suppose;D

 

This is a capacitor charging circuit. There is current flow when the capacitor is being charged.

 

The resistor in this circuit will have current flowing through it...

I don't get the point you're trying to make here.

 

That is incorrect. Let us begin the analysis:

First the current I:

Qualitative - since the circuit is opened there is no path for current therefore I=0.

Qualitative - using Ohms Law I=Vo/Ro, for an open circuit as Ro approaches infinity I approaches 0, therefore again I=0.

Now for voltage Vo:

Using Kirchhoff's Voltage law we can sum the voltages giving

-Vs + Vr + Vo = 0 or Vo = Vs - Vr

where Vr is the voltage across resistor R which can be defined by Ohm's Law as Vr = I*R. We have determined that I=0 therefore

Vr = 0 and Vo = Vs

In summary I = 0 and Vo = Vs

 

The current in a capacitor is defined as:

i = C dv/dt

where C is the capacitance and dv/dt is the time rate of change of the voltage.

In a DC voltage source circuit under steady state conditions the time rate of change of voltage is 0, therefore

dv/dt = 0 and i=0.

With i=0 we can consider the capacitor as an open circuit, which we demonstrated above. The capacitor voltage Vc = Vdc.

Again in summary i=0 and Vc=Vdc.

 

Yes, but in this example, the current and voltage are both zero, are they not?

 

No.

Current is zero and output voltage is equal to the input voltage.

For example the standard outlet in your house with nothing plugged in (open circuit) will have zero current and output voltage equal to input voltage of 120 VAC.

Or the cigarette lighter in your car with nothing plugged in (open circuit) will have zero current and output voltage equal to input voltage of 12 VDC.

 

still, you haven't explained how to solve for V=IR. If there is zero current, how do you solve V=IR?