Question

Obtain the equivalent capacitance of the network in Figure. For a 300 V supply, determine the charge and voltage across each capacitor.

Answer 1

Capacitance of capacitor C₁ is 100 pF.

Capacitance of capacitor C₂ is 200 pF.

Capacitance of capacitor C₃ is 200 pF.

Capacitance of capacitor C₄ is 100 pF.

Supply potential, V = 300 V

Capacitors C₂ and C₃ are connected in series.

Let their equivalent capacitance be C'

∴ `1/"C'" = 1/200 + 1/200 = 2/200`

∴ C' = 100 pF

Capacitors C₁ and C' are in parallel. Let their equivalent capacitance be C''.

∴ `"C''" = "C'" + "C"_1`

= 100 + 100

= 200 pF

C'' and C₄ are connected in series. Let their equivalent capacitance be C.

∴ `1/C = 1/("C''") + 1/("C"_4)`

= `1/200 + 1/100`

= `(2 + 1)/200`

C = `200/3` pF

Hence, the equivalent capacitance of the circuit is `200/3` pF.

Potential difference across C" = V"

Potential difference across C₄ = V₄

∴ `"V''" + "V"_4 = "V" = 300  "V"`

Charge on C₄ is given by

Q₄ = CV

= `200/3 xx 10^-12 xx 300`

= `2 xx 10^-8  "C"`

∴ `"V"_4 = "Q"_4/"C"_4`

= `(2 xx 10^-8)/(100 xx 10^-12)` = 200 V

∴ Voltage across C₁ is given by

`"V"_1 = "V" - "V"_4`

= `300 - 200 = 100  "V"`

Hence, potential difference, V₁, across C₁ is 100 V.

Charge on C₁ is given by,

`"Q"_1 = "C"_1"V"_1`

= `100 xx 10^-12 xx 100`

= `10^-8  "C"`

C₂ and C₃ have the same capacitances have a potential difference of 100 V together. Since C₂ and C₃ are in series, the potential difference across C₂ and C₃ is given by,

V₂ = V₃ = 50 V

Therefore, charge on C₂ is given by,

`"Q"_2 = "C"_2"V"_2`

= `200 xx 10^-12 xx 50`

= `10^-8  "C"`

And charge on C_{3 ­}is given by,

`"Q"_3 = "C"_3"V"_3`

= `200 xx 10^-12 xx 50`

= `10^-8  "C"`

Hence, the equivalent capacitance of the given circuit is `200/3` pF with 

Q₁ = 10⁻⁸ C, V₁ = 100 V

Q₂ = 10⁻⁸ C, V₂ = 50 V

Q₃ = 10⁻⁸ C, V₃ = 50 V

Q₄ = 2 × 10⁻⁸ C, V₄ = 200 V