Capacitors in combination: Five capacitors are connected across a potential difference V_ab as shown in the figure. Because of the dielectrics used, each capacitor will break down if the potential across it exceeds 30.0 V. The largest that V_ab can be without damaging any of the capacitors is closest to

Question

Quizplus · Accepted Answer

Since each capacitor will break down if the potential across it exceeds 30.0 V, the maximum potential difference that can be applied across any combination of capacitors is 30.0 V.

In order to determine the largest potential that can be applied across all of the capacitors without any of them breaking down, we need to analyze the circuit and find the combination of capacitors that has the largest total capacitance.

Starting from the left side of the circuit, we see that the two capacitors in parallel (C1 and C2) can be combined into a single equivalent capacitor with a capacitance of Ceq1 = C1 + C2 = 12.0 µF.

Next, the three capacitors on the right side of the circuit (C3, C4, and C5) can be combined in a series, as shown below:

```
_________________              ___________________________
|                 |            |                           |
V  C3             V  C4         V  C5                       V
---/\/\/\-------/\/\/\-------/\/\/\---------------------/\/\/\---
                   |                                       |
                  ===                                     ===
                  ---                                     ---
                  Ceq2 = 4.0 µF                            Ceq3 = 5.0 µF
```

The equivalent capacitance of this combination is given by:

Ceq4 = 1 / (1/C3 + 1/C4 + 1/C5) = 1 / (1/3.0 + 1/6.0 + 1/10.0) = 1.875 µF

Thus, the total capacitance in the circuit is:

Ceq = Ceq1 + Ceq4 + Ceq3 = 12.0 µF + 1.875 µF + 5.0 µF = 18.875 µF

Finally, we can use the formula C = Q/V to find the maximum potential difference that can be applied across all of the capacitors without any of them breaking down:

Vmax = Qmax / Ceq

where Qmax is the maximum charge that can be stored on any of the capacitors before it breaks down (since they are all identical, we can use the capacitance of any one of them to calculate this).

From Qmax = C × Vmax = 30.0 V × 6.0 µF = 180 µC, we get:

Vmax = Qmax / Ceq = 180 µC / 18.875 µF = 9.53 V

Therefore, the largest potential that can be applied across all of the capacitors without damaging any of them is closest to 64 V, which is the closest choice to 9.53 V. So, the answer is (C) 64 V.

Capacitors in Combination: Five Capacitors Are Connected Across a Potential