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Question

Find the minimum number of capacitors of 2 μf each required. To obtain a capacitance of 5 μf.


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Solution

Step1: Formula used

  1. When two capacitors are in series then their equivalent capacitance formula is 1Ceq=1C1+1C2=2μf2=1μf
  2. When two capacitors are in parallel then their equivalent capacitance is Ceq=C1+C2=2+2=4μf

Step2: Calculating the minimum number of capacitor

  1. If two capacitors(ABand BC) are placed in series and two capacitors (DE and GF) are placed parallel.
  2. We need to obtain equivalent captitance of 5μf. Therfore
  3. Demonstrating with the help of diagram

Case1: Capacitor AB and BC are in series.

  1. Let us assume {X} will be the equivalent capacitance of the capacitor connected in series.
  2. In series equivalent capacitance formula is 1Ceq=1C1+1C2=2μf2=1μf

Case2: Capacitor DE and GF are in parallel.

  1. Let us assume {Y} will be the equivalent capacitance of the capacitor connected parallelly.
  2. In parallel equivalent capacitance is Ceq=C1+C2=2+2=4μf

Case3: If capacitor Xand Y are in parallel then their equivalent capacitance becomes

Ceq=4+1=5 microfarad.

Hence, the minimum number of capacitors is 4.


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