Circuit Analysis - Resistor and Capacitor in parallel

[ratatat]

It says in the solution that the potential difference across the capacitor is 6V. I don't understand, because I thought the voltage in all parallel circuits was equal to the total voltage. ie. I figured the voltage would be 12V in both parallel circuits.

Question http://i31.tinypic.com/2vs52ts.jpg

Solution http://i29.tinypic.com/2hh26tj.jpg

Thanks for your help

---

Members don't see this ad.

 

[sleepy425]

It says in the solution that the potential difference across the capacitor is 6V. I don't understand, because I thought the voltage in all parallel circuits was equal to the total voltage. ie. I figured the voltage would be 12V in both parallel circuits.

Question http://i31.tinypic.com/2vs52ts.jpg

Solution http://i29.tinypic.com/2hh26tj.jpg

Thanks for your help

you would be correct if that second resistor were not in the circuit. If it were just a capacitor and a resistor in series, the voltage drop across each would have to be 12 V. However, that second resistor changes things. So here's how you figure it out:

When the circuit has been closed for a long time, the capacitor is almost fully charged. A fully charged capacitor is effectively an open circuit--no current flows through that part of the circuit. Soooo, you can simplify the circuit by ignoring the capacitor part, and you're left with 2 resistors in series. Since each resistor is 2 ohms, the total resistance is 4 ohms, so the current through that part of the circuit is 12 V/4 ohms=3 amps. So now that you know the current through that part of the circuit, you can find the voltage drop across the resistor that is in parallel with the capacitor: 3amps*2ohms=6V. Since that resistor is in parallel with the capacitor, the voltage drop across the capacitor must be the same as the voltage drop across the resistor, so it's 6 V. (Incidentally, instead of taking 12V/4 ohms to find the current and then multiplying back by 2 ohms, you could just recognize that you have a series circuit with 2 identical resistors, so the voltage drop should be the same across each resistor, so each resistor has half of the total voltage drop, so 6 V).

 

[ratatat]

Thanks a lot. That had been boggling my mind for quite some time. Much appreciated.

 

[Hemichordate]

If it were just a capacitor and a resistor in series, the voltage drop across each would have to be 12 V. However, that second resistor changes things. So here's how you figure it out:

But I thought in a series circuit, only the current stayed the same throughout, not the voltage?

 

[wanderer]

If the capacitor and resistor were in series, wouldn't it effectively be an open circuit?

 

[sleepy425]

But I thought in a series circuit, only the current stayed the same throughout, not the voltage?

oops, typo. I meant to write parallel.

wanderer, yes, if the capacitor and resistor were in series, at t=infinity, it would effectively be an open circuit. at t=0, it would be as though the capacitor were just a short circuit (in other words, it would be as though you just had the resistor but no capacitor in the circuit).

 

[ratatat]

Cool, thanks guys.