What Is the Breakdown Voltage of the Capacitor Combination?

[lizzyb]

Question:

Each capacitor in the combination shown in Figure P26.49 has a breakdown voltage of 15.0 V. What is the breakdown voltage of the combination?

http://img244.imageshack.us/img244/6705/serway5thp2649ue8.th.jpg

Work Done:

I figured that the voltage on the two parallel capacitors on either side must be the same, so both can handle 15V. So I assumed that, since C = Q/V, the amount of charge to the central capacitor would be 3 X 10^-4, but since the whole thing is serial, we have: V = V_1 + V_2 + V_3. How should I go about setting this one up?

 

[Kurdt]

What you need to do is work out the capacitance of the combination. What equations do you know that will relate the capacitance to the voltage equation you have written?

 

[lizzyb]

I can easily determine the entire capicitance of the whole system.
Let C_1 = 20 mu-F, C_2 = 10 mu-F, then
$$\frac{1}{C} = \frac{1}{2C_1} + \frac{1}{C_2} + \frac{1}{2C_1}$$
or $$C = \frac{4 C_1 C_2}{4C_1 + 2}$$

We also know the charge of the center capacitor if V = 15, and a serial circuit has all the same charge, so using the highest charge on the center capicitor and the above equation for the entire circuit,

$$V = \frac{Q}{C} = \frac{Q_c}{\frac{4 C_1 C_2}{4 C_1 + C_2}} = \frac{Q_c(4 C_1 + C2)}{4 C_1 C_2}$$

that's not it - I plugged in the numbers and it's not correct.

 

[Kurdt]

Ok I think you have the right idea but you've made a couple of mistakes.

Firstly $$\frac{1}{C} = \frac{1}{2C_1} + \frac{1}{C_2} + \frac{1}{2C_1} = \frac{4C_1 + 2}{4 C_1 C_2}$$

is not correct.

$$\frac{1}{C_{eff}} = \frac{1}{2C_1} + \frac{1}{C_2} + \frac{1}{2C_1} = \frac{1}{C_1} + \frac{1}{C_2}$$ that should get you started on getting the right fraction.

For the second part like I say you have the correct method. So for the middle capacitor which you have labelled C₂, the charge it stores at maximum voltage is:

$$\Delta V_{max}=\frac{Q}{C_2}\Rightarrow Q = \Delta V_{max}C_2$$

now substitute Q into the $$\Delta V_{max}=\frac{Q}{C_{eff}}$$ equation and see how you go from there.

 

[lizzyb]

Yes! thanks ever much! :-)