Two capacitors of 1 mu F and 2 mu F are con | vedclass

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

Question

(C) In a series combination,the charge $Q$ on each capacitor is the same.The equivalent capacitance $C_{eq}$ is given by:$C_{eq} = \frac{C_1 C_2}{C_1

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(C) In a series combination,the charge $Q$ on each capacitor is the same.The equivalent capacitance $C_{eq}$ is given by:$C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{1 	imes 2}{1 + 2} \ \mu F = \frac{2}{3} \ \mu F$The total charge $Q$ stored in the combination is:$Q = C_{eq} V = \left( \frac{2}{3} 	imes 10^{-6} \ F ight) 	imes 120 \ V = 80 	imes 10^{-6} \ C$The potential difference $V_1$ across the $1 \ \mu F$ capacitor is:$V_1 = \frac{Q}{C_1} = \frac{80 	imes 10^{-6} \ C}{1 	imes 10^{-6} \ F} = 80 \ V$

Two capacitors of $1 \ \mu F$ and $2 \ \mu F$ are connected in series. This combination is charged to a potential difference of $120 \ V$. What is the potential difference across the $1 \ \mu F$ capacitor in volts?

Similar Questions

Three capacitors each of capacitance $C$ and breakdown voltage $V$ are connected in series. The capacitance and breakdown voltage of the series combination will be respectively:

Three capacitors of capacitances $C_1=2 \mu F$,$C_2=3 \mu F$ and $C_3=5 \mu F$ are connected in series. $A$ potential difference of $155 \ V$ is applied across the combination. Choose the correct option.

The potential difference across the $4 \mu F$ capacitor in series with the parallel combination in the following circuit is: (in $V$)

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

If three capacitors each of capacity $1\,\mu F$ are connected in such a way that the resultant capacity is $1.5\,\mu F$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module