Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

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
  • 1 month ago

    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

Still have questions? Get your answers by asking now.