Grounding a circuit through a capacitor/parallel RC circuit

[jozefmak]

Hi,
I have two similar circuits. One of them is clear for me but I am not sure how the second one works.

Let's start with the clear one:

a parallel RC circuit, which is explained here:

At time t = 0s there are zero volts across the resistor because the capacitor is fully uncharged. After some time, the capacitor will get charged to 12V so we will measure 12V across the resistor.

This is the unclear circuit for me:

Will the capacitor get charged? (What voltage?)
What voltage are we going to measure at t = 0s across the resistor?

I know that when I remove the capacitor:

there will be 12V across the resistor at time t = 0s because a positive terminal of battery will have 0V and negative terminal of battery will have -12V (with respect to those 0V). The problem, which makes this circuit unclear for me is the capacitor.

 

[phyzguy]

In your second circuit, there will be 12V across the resistor. But the voltage at the top of the capacitor is unknown, and could be anything (within limits set by the breakdown voltage of the capacitor). So the voltages at the resistor terminals could be +10V and -2V, or +100V and +88V, or -10V and -22V, or anything you like. The voltage that actually appears depends on the past history of the circuit.

 

[jozefmak]

I see, thanks for reply!

I am just wondering for better understanding what exactly this sentence means:
"The voltage that actually appears depends on the past history of the circuit."
Can you give some example, please?

 

[Dale]

Where are you connecting the 9V battery?

 

[phyzguy]

Suppose you take the capacitor and connect a wire to both sides and discharge it. Then there is no voltage across it. So when you connect the bottom side to 0V, the top is at 0V as well. So when you connect the battery and resistor to it, the top of the resistor will be at 0V and the bottom will be at -12V. On the other hand, suppose you connect the capacitor to a power supply and charge it up to 100V. Then when you connect the bottom of the capacitor to 0V, the top will be at +100V. So when you connect the battery and resistor to it, the top of the resistor will be at +100V and the bottom will be at +88V. All of this assumes an ideal capacitor, and that the capacitance of the capacitor is large compared to the capacitance of the battery and resistor. In a real circuit, there will be charge sharing and it gets more complicated.

 

[sophiecentaur]

It's a sort of hackneyed phrase but you really need a CIRCUIT with a path between the connections of any component. The 'Unclear' circuit is unclear because there are not enough connecting wires.

 

[Dale]

The 'Unclear' circuit is unclear because there are not enough connecting wires.

Good point. The deficiency is not in the OP’s knowledge. It is a deficiency in the circuit.

 

[jozefmak]

Suppose you take the capacitor and connect a wire to both sides and discharge it. Then there is no voltage across it. So when you connect the bottom side to 0V, the top is at 0V as well. So when you connect the battery and resistor to it, the top of the resistor will be at 0V and the bottom will be at -12V. On the other hand, suppose you connect the capacitor to a power supply and charge it up to 100V. Then when you connect the bottom of the capacitor to 0V, the top will be at +100V. So when you connect the battery and resistor to it, the top of the resistor will be at +100V and the bottom will be at +88V.

OK, now I understand. Voltage across the capacitor will be invariable (no charging, no discharging) when connected in the "unclear" scheme. Thank you!