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# Algebra-based physics question?

Suppose you have a large bag of 4.0 nF capacitors and ample wire to connect them.

1. Find a configuration of capacitors from your bag with wires connecting them that has the following equivalent capacitance:

a. 10 nF

b. 3 nF

2. Find a way to connect 4 of the capacitors from your bag to get an equivalent capacitance of 4.0 nF. [All 4 capacitors should be in the circuit.]

3. For all of the configurations in parts 1 and 2, find the potential difference of a battery needed to store a total of 75 nJ of energy.

4. For each of the configurations in part 1, find the potential difference of a battery across the configuration such that the maximum amount of charge on any of the capacitors is exactly 24 nC. [In each case there may be more than one capacitor with that amount of charge.]

### 1 Answer

Relevance
• 1a) a pair in series is 2 nF. 5 such pairs in parallel is 10

there are other answers....

b) 4 units in series is 1 nF. 3 such sets in parallel is 3 nF

there are other answers....

2) two in parallel = 8, two such sets in series = 4

or

two in series = 2, two such sets in parallel = 4

3)

E = ½CV²

75 nJ = ½10nF • V²

V² = 15

V = 3.87 volts

E = ½CV²75 nJ = ½3nF • V²V² = 50V = 7.07 volts

d) depends on the configuration used in a and b

a pair in series is 2 nF. 5 such pairs in parallel is 10

take one cap, 24 nC = ½4nF•V²

V² = 48, V = 6.93 volts

two in series, total V = 13.9 volts

4 units in series is 1 nF. 3 such sets in parallel is 3 nF

take one cap, 24 nC = ½4nF•V²V² = 48, V = 6.93 volts4 in series, total V = 27.7 volts

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