Question

Solve this

9. Two capacitances of capacity C₁ and C₂ connected in series and potential difference V is applied across it. Then the potential difference across C₁ will be [MP PMT 1985]
$\left(a\right)V\frac{C_2}{C_1}\left(b\right)V\frac{C_{1+}{C}_2}{C_1}\left(c\right)V\frac{C_2}{C_1+C_2}\left(d\right)V\frac{C_1}{C_1+C_2}$

Answer 1

Dear student
$$\begin{gathered} Equivalentcapaci\tan ceofthecapacitorsis \\ {C}_{eqv}=\frac{C_1{C}_2}{C_1+C_2} \\ Chargesuppliedbythecellis \\ q=C_{eqv}\times V=\frac{C_1{C}_2}{C_1+C_2}V \\ chargeonboththecapacitorsisthesameandisequalto=\frac{C_1{C}_2}{C_1+C_2}V \\ potentialdifferenceacrossthecapacitorofcapaci\tan ce{C}_{1}is \\ {V}_1=\frac{q}{C_1}=\frac{C_1{C}_2}{C_1(C_1+C_2)}V=\frac{C_2}{C_1+C_2}V \\ Regards \end{gathered}$$