Four identical capacitors are connected in se | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $q$ be the charge on each capacitor. Since they are in series,the charge $q$ is the same for all.The total potential difference across the fou

Vedclass · Accepted Answer

$7.5$

Vedclass · Answer

(B) Let $q$ be the charge on each capacitor. Since they are in series,the charge $q$ is the same for all.The total potential difference across the four capacitors is $10 \, V$. Therefore,$\frac{q}{C} + \frac{q}{C} + \frac{q}{C} + \frac{q}{C} = 10 \, V$,which gives $\frac{q}{C} = 2.5 \, V$.Point $X$ is earthed,so its potential $V_X = 0 \, V$.Looking at the circuit diagram,point $A$ is connected to the positive terminal of the battery,and there are three capacitors between point $A$ and point $X$. The potential difference between $A$ and $X$ is $V_A - V_X = \frac{q}{C} + \frac{q}{C} + \frac{q}{C} = 3 	imes 2.5 \, V = 7.5 \, V$.Since $V_X = 0 \, V$,we have $V_A = 7.5 \, V$.

Four identical capacitors are connected in series with a battery of $emf$ $10 \, V$. The point $X$ is earthed. Find the potential of point $A$. (in $, V$)

Similar Questions

Assertion : If three capacitors of capacitances $C_1 < C_2 < C_3$ are connected in parallel,then their equivalent capacitance $C_P > C_S$,where $C_S$ is the equivalent capacitance in series.
Reason : $\frac{1}{C_P} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

The capacitance between the points $A$ and $B$ in the given circuit will be.......$\mu F$.

The equivalent capacitance between points $A$ and $B$ in the figure shown below is .......... $\mu F$.

Four identical condensers are connected in parallel and then in series. The ratio of equivalent capacitance in series to that in parallel combination is

The equivalent capacitance between terminals $A$ and $B$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module