How Can a Capacitor be Charged to the Voltage of the Battery in an RC Circuit?

[kfreshn]

Resistor Capacitance Circuits.Say there is a resistor before a capacitor connected in a wired circuit; and there is a battery as the emf source (10V for example).

How can the capacitor be charged to approx. the voltage of the battery if the voltage drop across the resistor = the voltage of the battery (V=IR and Kirschoff's Law states the voltage drop across the resistor should equal the Voltage of the battery)?

I guess I'm having trouble understanding how a resistor can reduce voltage of the circuit to 0, and yet current is still able to flow after it leaves the resistor --- albeit at a slower rate?.

I thought of Voltage = 0, there would be no more current movement. Need explanation on this.

And how can the Capacitor be charged to the voltage of the battery. If voltage drop across resistor is 10V, how can the capacitor be charged to 10V?

ugh.

Thanks for all the help in advance.

I'll Forever grateful for anyone who can help me asap :)

 

[berkeman]

Resistor Capacitance Circuits.


Say there is a resistor before a capacitor connected in a wired circuit; and there is a battery as the emf source (10V for example).

How can the capacitor be charged to approx. the voltage of the battery if the voltage drop across the resistor = the voltage of the battery (V=IR and Kirschoff's Law states the voltage drop across the resistor should equal the Voltage of the battery)?

I guess I'm having trouble understanding how a resistor can reduce voltage of the circuit to 0, and yet current is still able to flow after it leaves the resistor --- albeit at a slower rate?.

I thought of Voltage = 0, there would be no more current movement. Need explanation on this.

And how can the Capacitor be charged to the voltage of the battery. If voltage drop across resistor is 10V, how can the capacitor be charged to 10V?

ugh.

Thanks for all the help in advance.

I'll Forever grateful for anyone who can help me asap :)

Welcome to the PF.

(I moved your thread to here in the Homework Help section of the forums, where schoolwork questions like this should be posted.)

Are you familiar with the differential equation that relates v(t) and i(t) for a capacitor? That's really the key to understanding how this circuit behaves. Can you show us that equation?

 

[kfreshn]

IR(t) + q(t)/C = V(t)?

edit: thanks for the welcome

 

[cepheid]

harged to approx. the voltage of the battery if the voltage drop across the resistor = the voltage of the battery (V=IR and Kirschoff's Law states the voltage drop across the resistor should equal the Voltage of the battery)?

No, Kirchoff's law says the sum of the voltages around a closed loop should be zero, so the resistor voltage drop PLUS the capacitor voltage drop must add up to the battery voltage. At the end of the charging process, the current goes to zero, which means there is no voltage drop across the resistor, and all of the battery voltage has developed across the capacitor.

Also: what berkeman said.

 

[kfreshn]

So will the resistor in the RC circuit affect the electric field between the capacitor? V=ED?

Ty for your help.

 

[jhae2.718]

IR(t) + q(t)/C = V(t)?

edit: thanks for the welcome

To describe the charging of an RC circuit, you'll need to solve this ODE. First, you need to get it in terms of one variable. How can you write i in terms of q?
$$Ri(t) + \frac{1}{C}q(t) = V(t)$$

Note that the solution to this equation is wrong. To accurately describe the circuit, you'll need to take self-inductance into account, which will give you an RLC circuit, which you should get to soon.