EK Physics Capacitor Question

[Postictal Raiden]

---

Members don't see this ad.

In the 30-min exam for Lecture 7, question #159 asks about the the charge of the capacitor in a circuit. Since the Voltage of the battery is 12V, and the charge of the capacitance is 1uF, shouldn't the charge on the capacitor, after it had been fully charged, = 12V x 1uF = 1.2x10^-5? The answer is half of that (6 x10^-6).

 

[hiaips]

In the 30-min exam for Lecture 7, question #159 asks about the the charge of the capacitor in a circuit. Since the Voltage of the battery is 12V, and the charge of the capacitance is 1uF, shouldn't the charge on the capacitor, after it had been fully charged, = 12V x 1uF = 1.2x10^-5? The answer is half of that (6 x10^-6).

The voltage across the capacitor is 6 V instead of 12 V.

To see this, first write down Kirchoff's Law for the circuit loop consisting of two resistors and note that the voltage drop across each resistor is 6 V. Since parallel branches of a circuit have the same voltage drop across them, the voltage drop across the capacitor must also be 6 V.

 

[Postictal Raiden]

The voltage across the capacitor is 6 V instead of 12 V.

To see this, first write down Kirchoff's Law for the circuit loop consisting of two resistors and note that the voltage drop across each resistor is 6 V. Since parallel branches of a circuit have the same voltage drop across them, the voltage drop across the capacitor must also be 6 V.

OK, I understand that to find the voltage drop across each resistor, I first need to find the Current. So, if Reff = (2 resistors in series, each has a value of 2ohms), I= 12V/4ohms = 3A. Here, I get stuck. I don't know how to go from knowing that across each resistor the voltage is 6 to the fact that the voltage across the capacitor is also 6. Shouldn't all voltage values in the cell add up to the voltage of the batter?

Sorry if I'm sounding annoyingly stupid, but I really suck at this.

 

[hiaips]

OK, I understand that to find the voltage drop across each resistor, I first need to find the Current. So, if Reff = (2 resistors in series, each has a value of 2ohms), I= 12V/4ohms = 3A. Here, I get stuck. I don't know how to go from knowing that across each resistor the voltage is 6 to the fact that the voltage across the capacitor is also 6. Shouldn't all voltage values in the cell add up to the voltage of the batter?

Sorry if I'm sounding annoyingly stupid, but I really suck at this.

Not quite. Using Kirchoff's Law, the voltage drops around a series circuit (or a single loop) will add up to 0. However, we have two loops here. The sum of voltages in both loops must add up to 0 separately and simultaneously.

From analyzing the loop containing the two resistors, you can figure out that the voltage drop across each resistor is 6V. Now look at the loop containing the capacitor. It also passes through one of the resistors, right? Let's analyze that loop using Kirchoff's law:

Vb - Vc - Vr = 0

But we know that Vb = 12V (given) and Vr = 6V (calculated from analyzing the other circuit loop); hence,

12 - Vc - 6 = 0.

So, Vc must be 6V.

 

[Postictal Raiden]

Not quite. Using Kirchoff's Law, the voltage drops around a series circuit (or a single loop) will add up to 0. However, we have two loops here. The sum of voltages in both loops must add up to 0 separately and simultaneously.

From analyzing the loop containing the two resistors, you can figure out that the voltage drop across each resistor is 6V. Now look at the loop containing the capacitor. It also passes through one of the resistors, right? Let's analyze that loop using Kirchoff's law:

Vb - Vc - Vr = 0

But we know that Vb = 12V (given) and Vr = 6V (calculated from analyzing the other circuit loop); hence,

12 - Vc - 6 = 0.

So, Vc must be 6V.

I think I got it now.

So, the key point was to have a total voltage in each loop equals to zero. The first loop has two resistors of identical magnitude so each should have voltage equals to 6 to cancel out with the batter voltage. The second loop has one of the resisters of the first loop (voltage=6) and a capacitor which now acts as a battery. To have a net voltage value of zero in the second loop, the voltage across the capacitor and that of the resistor must be equal, must be 6V.

Thank you.