Frequently, several capacitors are connected together to form a collection of capacitors. We may be interested in determining the overall capacitance of such a collection. The simplest configuration to analyze involves capacitors connected in series or in parallel. More complicated setups can often (though not always!) be treated by combining the rules for these two cases. Consider the example of a parallel combination of capacitors: Three capacitors are connected to each other and to a battery as shown in the figure. The individual capacitances are C, 2C, and 3C, and the battery’s voltage is V.

If the potential of plate 1 is V, then, in equilibrium, what are the potentials of plates 3 and 6? Assume that the negative terminal of the battery is at zero potential.

If the charge of the first capacitor (the one with capacitance C) is Q, then what are the charges of the second and third capacitors?

Suppose we consider the system of the three capacitors as a single "equivalent" capacitor. Given the charges of the three individual capacitors calculated in the previous part, find the total charge Qtot for this equivalent capacitor.

Using the value of Qtot, find the equivalent capacitance Ceq for this combination of capacitors.

—1–C–2—

–3–2C-4–

–5-3C–6–

—-V——

### 3 Answers

as shown in the figure ??

"Consider the example of a parallel combination of capacitors: Three capacitors are connected to each other and to a battery as shown in the figure. The individual capacitances are C, 2C, and 3C, and the battery’s voltage is V."

OK, we have 3 parallel capacitors, you seem to be saying.

"If the potential of plate 1 is V, then, in equilibrium, what are the potentials of plates 3 and 6?"

?? you have 3 caps, where did plate 6 come from? In any case, as they are in parallel, all three have the same voltage.

"If the charge of the first capacitor (the one with capacitance C) is Q, then what are the charges of the second and third capacitors?"

2Q and 3Q

"find the total charge Qtot for this equivalent capacitor."

3+2+1 = 6

last part makes even less sense than the first parts.

Since 1 and 3 are connected, the potential of 1 is the same as the potential of 3 and = V,

Similarly since 2, 4, and 6 are connected to each other and to the negative terminal of the battery their potential is the same as that of the negative terminal of the battery that is zero.

Since C = Q/V or Q = CV

The charge on the first capacitor is C V

that of the second is 2CV and that of the third is 3CV

Total charge is Q total = C V+ 2CV + 3CV = 6 CV

The equivalent capacitance C eq = Q total / V

= 6CV/V = 6C

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Doubling the plate separation will halve the capacitance. this suggests you elect two times the voltage to shop the comparable fee. the comparable fee is latest on the capacitor because of the fact it relatively is not reconnected to something, or its plates shorted out. possibly the hot voltage would be 20V? examine this because of the fact there is not any longer a lot determining, only an comprehend-how of capacitors and capacitance. The formula for the capacitance of a parallel plate capacitor has ‘t’ (plate separation) on the base line for this reason doubling it halves the capacitance and why great fee capacitors have plates close at the same time.