Gujarati
Hindi
1. Electric Charges and Fields
normal

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

A

$10$

B

$7.5$

C

$-7.5$

D

$0$

Solution

If $\mathrm{q}$ be the change on each capacitor, then

$\frac{q}{C}+\frac{q}{C}+\frac{q}{C}+\frac{q}{C}=10 \mathrm{\,V} \Rightarrow \frac{q}{C}=2.5 \mathrm{\,V}$

$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{x}}=\frac{\mathrm{q}}{\mathrm{C}}+\frac{\mathrm{q}}{\mathrm{C}}+\frac{\mathrm{q}}{\mathrm{C}}=7.5 \mathrm{\,V}$

Standard 12
Physics

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