Combination of Capacitors: Overview, Questions, Preparation

A

Capacitance of each of the three capacitors, C = 9 pF

• The equivalent capacitance when the capacitors are connected in series is given by

 =  +  =  =  =

Hence,  = 3 pF

• Supply voltage, V = 120 V

Potential difference ( ) across each capacitor is given by  =  =  = 40 V

A. 

Let the three capacitors be

= 2 pF,  = 3 pF and  = 4 pF

• The equivalent capacitance, is given by

 = 2 + 3 + 4 = 9 pF

  When the combination is connected to a 100 V supply, the voltage in all three capacitors = 100 V

charge in each capacitor is given by the relation, q = VC

Hence for ,  = 100  = 200 pC = 2 C,

for ,  = 100  = 300 pC = 3 C,

for ,  = 100  = 400 pC = 4 C

A. Area of each plate, A = 6  

Distance between plates, d = 3 mm = 3  m

Supply voltage, V = 100 V

 

Capacitance C of the parallel plate is given by C =

In case of air, dielectric constant k = 1 and

 = permittivity of free space =  8.854    

Hence C =  F = 17.708 F = 17.71 pF

Charge on each plate of the capacitor is given by q = VC = 100  C

= 1.771  C

Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is 1.77  C

A. Dielectric constant of the mica sheet, k =6

• While the voltage supply remained connected :

V = 100 V

Initial capacitance, C = 17.708 F

New capacitance,  = kC = 6 17.708  = 106.25  F

= 106.25 pF

New charge,  =  = 106.25  C = 10.62  C

• If the supply voltage is removed, then there will be constant amount of charge in the plates.

Charge, q = CV = 17.708  C = 1.7708  C

Potential across plates,  =  =  = 16.66 V

A. Let C’ be the equivalent capacitance for capacitors  and  connected in series.

Hence,  =  + . So C’ = 100 pF

Capacitors C’ and  are in parallel, if the equivalent capacitance be C”, then

C” = C’ +  = 100 + 100 = 200 pF

Now C” and are connected in series. If the total equivalent capacitance of the circuit be , then  =  +  =  + ,   =  pF =  F

Let V” be the potential difference across C” and  be the potential difference across

Then, V” +  = 300 V

Now, charge on is given by =  =   = 2 C

 =  = 200. So V” = 100 V

So potential difference across C’ and  is V” = 100 V

Charge on  is given by = 100  C = 1  C

 and  are having same capacitance and have a potential difference of 100V together. Since  and  are in series, the potential difference is given by  =  = 50V

Hence the charge on  is given by  = 200  = C and the charge on  is given by  = 200  = C

Therefore, the equivalent capacitance of the circuit is  pF and charge and voltage at all capacitance is given as

For  =  C,  = 100 V

For  =  C,  = 50 V

For  =  C,  = 50 V

For  = 2 C,  = 200 V