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1. Electric Charges and Fields
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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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