Two capacitances of capacity C_1 and C_2 are | vedclass

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

Question

(C) When capacitors are connected in series,the charge $Q$ on each capacitor is the same.The equivalent capacitance $C_{eq}$ of the series combination

Vedclass · Accepted Answer

$V \frac{C_2}{C_1 + C_2}$

Vedclass · Answer

(C) When capacitors are connected in series,the charge $Q$ on each capacitor is the same.The equivalent capacitance $C_{eq}$ of the series combination is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{C_1 + C_2}{C_1 C_2}$,so $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$.The total charge $Q$ stored in the combination is $Q = C_{eq} V = \frac{C_1 C_2 V}{C_1 + C_2}$.The potential difference $V_1$ across capacitor $C_1$ is given by $V_1 = \frac{Q}{C_1}$.Substituting the value of $Q$,we get $V_1 = \frac{1}{C_1} 	imes \left( \frac{C_1 C_2 V}{C_1 + C_2} ight) = \frac{C_2 V}{C_1 + C_2}$.

Two capacitances of capacity $C_1$ and $C_2$ are connected in series and a potential difference $V$ is applied across the combination. The potential difference across $C_1$ will be:

Similar Questions

$A$ $5 \mu F$ capacitor is connected in series with a $10 \mu F$ capacitor. When a $300 \ V$ potential difference is applied across this combination,the total energy stored in the capacitors is (in $J$)

In the given figure,the equivalent capacitance $C_{AB}$ is equal to ...... $C$.

To obtain $3\,\mu F$ capacitance from three capacitors of $2\,\mu F$ each,they will be arranged as:

Seven capacitors,each of capacitance $2 \mu F$,are connected in a configuration to obtain an effective capacitance of $\frac{6}{13} \mu F$. The combination that will achieve this is:

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.

Vedclass Test Series

Exam Paper Generator

Online Exam Module