Question

Three capacitors each of capacitance 9 pF are connected in series (i) What is the total capacitance of the combination? (ii) What is the potential difference across each capacitor, if the combination is connected to 120 V supply?

Answer 1

Hey There!!


1) Here we are given that three capacitors, each 9 pF, are connected in series. 

In series connection of three capacitors, the equivalent capacitance is given by:

$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \\ \\ \\ \implies \frac{1}{C_{eq}} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} \\ \\ \\ \implies \frac{1}{C_{eq}} = \frac{3}{9} \\ \\ \\ \implies \frac{1}{C_{eq}} = \frac{1}{3} \\ \\ \\ \implies \boxed{C_{eq}=3 \, \, pF}$


2) Now, the combination is connected to a 120 V supply. 

In a series connection of capacitors, the charge across each capacitor is same, but the potential is divided among the capacitors according to their capacitances. 


Here, capacitance is same for each of the three capacitors. So, potential drop will also be same across each capacitor.

Total potential is 120 V. And there are three capacitors. 

So, potential drop across each capacitor would be $\frac{120}{3}=40$ V. 


Thus, Potential Drop across each capacitor is 40 V.




Hope it helps
Purva
Brainly Community

Answer 2

Given Three capacitors each of capacitance 9 pF are connected in series.
Let C1=C2=C3=9pF=9x10⁻¹²F

Let Cs be the effective capacity of series connection then,
1/CS= 1/C1+1/C2+1/C3
1/Cs=[1/9+1/9+1/9]1/10⁻¹²
1/Cs=[3/9]x1/10⁻¹²
1/Cs=1/3x10⁻¹² 
Cs=3x10⁻¹²F=3pF

Let v1, v2 qnd v3 be the potential differences across 3 capacitors 
v1+v2+v3=120
if q is the charge on each capacitor on each capacitor
q/C1+q/C2+q/C3=120
q[1/C1+1/C2+1/C3]=120
q=120xCs
q=120 x 3x10⁻¹² C
q=360x10⁻¹²C

Potential difference Across each capacitor = q/C
= 360x10⁻¹²/9x10⁻¹²=40 Volts