Question on capacitors in an RC circuit

[BlueCerealBox]

Homework Statement

All information is given in the picture posted

Homework Equations

V=IR , Q=CV

The Attempt at a Solution

What I don't understand is , when the capacitors are initially uncharged , they should just be treated as wires , if this is the case , then the entire circuit is shorted. So why would there be a potential at points a and b?

 

[STEMucator]

So why would there be a potential at points a and b?

Assume for a moment that the $24V$ source is shorted to zero (put a wire there) and the switch is open. So all you have is a circuit with some components and nothing going on. There is no potential at any point, and no current to be measured at all. This is reasonable because we have no power supply yet.

Now suppose we apply the $24V$ source to the circuit with the switch open. Immediately, current will begin to flow through the circuit and the capacitors will begin charging up. This implies $V = IR \neq 0$ and $V = \frac{q}{C} \neq 0$. So we can deduce there must be a potential at points $a$ and $b$. In fact, because the resistors and capacitors are in parallel, we know the $24V$ must drop across both branches.

 

[BlueCerealBox]

Assume for a moment that the $24V$ source is shorted to zero (put a wire there) and the switch is open. So all you have is a circuit with some components and nothing going on. There is no potential at any point, and no current to be measured at all. This is reasonable because we have no power supply yet.

Now suppose we apply the $24V$ source to the circuit with the switch open. Immediately, current will begin to flow through the circuit and the capacitors will begin charging up. This implies $V = IR \neq 0$ and $V = \frac{q}{C} \neq 0$. So we can deduce there must be a potential at points $a$ and $b$. In fact, because the resistors and capacitors are in parallel, we know the $24V$ must drop across both branches.

So it's only for this special case? If I were to put a resistor directly after the battery source , would the following case still apply?

 

[STEMucator]

So it's only for this special case? If I were to put a resistor directly after the battery source , would the following case still apply?

I don't believe there's anything too special about it.

If you were to place a resistor directly after the battery source (it could be the internal resistance of the battery for example) then the calculations would change somewhat. Some of the voltage would drop across this resistance. Call the amount of voltage dropped across this resistance $V_B$ for a moment.

Then, when we first apply the $24V$ source, we know the voltage remaining after the $V_B$ drop is $24V - V_B$.

This remaining $24V - V_B$ would then be lost across both parallel branches. We can measure the potential at points $a$ and $b$ by measuring the voltage across the $4.4 \Omega$ resistor and the $0.24 \mu F$ capacitor respectively.

This is assuming the switch is open at the time.

 

[BlueCerealBox]

Ah ok. I think I somewhat understand it now. Thanks for the help!