Question

Derive an expression for the effective capacitance of three parallel plate capacitors connected in series.

Answer 1

1. Diagram:
   
   
2. Explanation:  
   a. Capacitors are said to be connected in series if they are connected one after the other in the form of a chain.
   b. Let capacitors of capacitances C₁, C₂, C₃ be connected in series as shown in the figure.
   c. Let V₁, V₂, V₃ be the corresponding potential differences in the capacitors.
   d. Suppose a potential difference ‘V’ is applied across the combination. The left plate of capacitor C₁ has a charge +Q. An equal but opposite charge -Q is induced on the right plate of this capacitor. This charge -Q induces a charge +Q on the left plate of C₂ and so on.   
   e. Thus, each capacitor receives a magnitude of charge Q. Hence, when the capacitors are connected in series, the same current flows through them and all have the same charge +Q induced on the plate. Thus, potential difference induced across capacitors is given by,
   `"V"_1 = "Q"/"C"_1, "V"_2 = "Q"/"C"_2, "V"_3 = "Q"/"C"_3` .......(1)
   f. Total potential difference ‘V’ across the combination is given by,
   V = V₁ + V₂ + V₃
   ∴ V = `"Q"/"C"_1 + "Q"/"C"_2 + "Q"/"C"_3` ….[From equation (1)]
   ∴ V = `"Q"(1/"C"_1 + 1/"C"_2 + 1/"C"_3)` …(2)
   g. If these capacitors are replaced by a single capacitor of capacity C_S, such that effective capacity remains the same then
   `"C"_"s"` = `"Q"/"V"`
   ∴ V = `"Q"/"C"_"s"`….(3)
   From equation (2) and (3),
   `"Q"/"C"_"s" = "Q"(1/"C"_1 + 1/"C"_2 + 1/"C"_3)`
   ∴ `1/"C"_"s" = (1/"C"_1 + 1/"C"_2 + 1/"C"_3)`
   ∴ `"C"_"s" = ("C"_1"C"_2"C"_3)/("C"_1"C"_2 + "C"_2"C"_3 + "C"_3"C"_1)`