Question

Four capacitors are connected in series with a battery of emf 10 V as shown in figure. The points P is earthed. The potential of point A is equal in magnitude to potential of point B but opposite in sign if:

• A. $C_1+C_2+C_3=C_4$
• B. $\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{C_4}$
• C. $\frac{C_1{C}_2{C}_3}{C_1^2+C_2^2+C_3^2}=C_4$
• D. $C+4=(C_1{C}_2{C}_3{)}^{1/3}$

Answer 1

Answer: (B)

$\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{C_4}$

Here the eqivalent capacitance between A and P is $C_{AP}=(1/C_1+1/C_2+1/C_3{)}^{-1}$ and $C_{PB}=C_4$
As $C_{AP}$ and $C_{PB}$ are in series so charge on each is same.
thus, $Q_{AP}=Q_{PB}\Rightarrow C_{AP}{V}_{AP}=C_{PB}{V}_{PB}$
Here, $V_{AP}=V_{PB}$ so $C_{AP}=C_{PB}$
thus, $(1/C_1+1/C_2+1/C_3{)}^{-1}=C_4$
or $\frac{1}{C_4}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$