Question

Q.2 three capacitors each of capacitance 9 pf are connected in series, across a 120 v supply. the total capacitance of the combination and the potential difference across each capacitor are respecitvely :

Answer 1

Dear user !!

since all the 3 capacitors have equal capacitance & are connected in

series, thus:

C (net) = C/3

             = 9/3 = 3pF

Now, Cnet = 3pF

 Vnet = 120V

and we know that Q = CV

>>>Qnet = 360 X 10^-12   c   {since 1pF = 10^-12 F}


For each capacitor:

Q = 360X10^-12 C

C = 9X10^-12 F

thus V = Q/C

            = 40 V



Hope it helps you !!