(C) In a series combination of capacitors,the charge $Q$ on each capacitor is the same.The potential difference across a capacitor is given by $V = \f

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(C) In a series combination of capacitors,the charge $Q$ on each capacitor is the same.The potential difference across a capacitor is given by $V = \frac{Q}{C}$.From the graph,the potential drop across the first capacitor $(C_1)$ is greater than the potential drop across the second capacitor $(C_2)$.Let $V_1$ be the potential difference across $C_1$ and $V_2$ be the potential difference across $C_2$.Since $V_1 > V_2$ and $Q = C_1 V_1 = C_2 V_2$,we have $\frac{C_1}{C_2} = \frac{V_2}{V_1}$.Since $V_1 > V_2$,it follows that $\frac{V_2}{V_1} < 1$,which implies $\frac{C_1}{C_2} < 1$,or ${C_1} < {C_2}$.

Class 12 Physics Electric Potential and Capacitance Equivalent Capacitance of Capacitor connected in Series and Parallel

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Figure $(a)$ shows two capacitors connected in series and joined to a battery. The graph in figure $(b)$ shows the variation in potential as one moves from left to right on the branch containing the capacitors. If the graph represents the potential across the capacitors,then:

A
${C_1} > {C_2}$
B
${C_1} = {C_2}$
C
${C_1} < {C_2}$
D
The information is not sufficient to decide the relation between ${C_1}$ and ${C_2}$

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Electric Potential and Capacitance

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Equivalent Capacitance of Capacitor connected in Series and Parallel

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 : $\frac{1}{C_P} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

EasyAIIMS 2002

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$A$ parallel combination of two capacitors of capacities $2C$ and $C$ is connected across a $5 \text{ V}$ battery. When they are fully charged,the charges and energies stored in them are $Q_1, Q_2$ and $E_1, E_2$ respectively. Then $\frac{E_1-E_2}{Q_1-Q_2}$ in $\text{J/C}$ is (capacity is in Farad,charge in Coulomb and energy in $\text{J}$)

DifficultMHT CET 2023

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Find the equivalent capacitance of the system of capacitors between points $A$ and $B$ as shown in the figure.

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Ten capacitors are connected in parallel to a battery of $V$ volts. If all capacitors are disconnected from the battery and then connected in series,what will be the voltage across the combination in terms of $V$ (in $V$)?

Easy

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What is the equivalent capacitance between the points $P$ and $Q$ in the combination of capacitors shown in the figure (in $\mu F$)?

EasyMHT CET 2020

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Figure $(a)$ shows two capacitors connected in series and joined to a battery. The graph in figure $(b)$ shows the variation in potential as one moves from left to right on the branch containing the capacitors. If the graph represents the potential across the capacitors,then:

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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 : $\frac{1}{C_P} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

Find the equivalent capacitance of the system of capacitors between points $A$ and $B$ as shown in the figure.

Ten capacitors are connected in parallel to a battery of $V$ volts. If all capacitors are disconnected from the battery and then connected in series,what will be the voltage across the combination in terms of $V$ (in $V$)?

What is the equivalent capacitance between the points $P$ and $Q$ in the combination of capacitors shown in the figure (in $\mu F$)?

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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 : $\frac{1}{C_P} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$