Four capacitors with capacitances $C_1 = 1\,\mu F, C_2 = 1.5\,\mu F, C_3 = 2.5\,\mu F$ and $C_4 = 0.5\,\mu F$ are connected as shown and are connected to a $30\, volt$ source. The potential difference between points $a$ and $b$ is.....$V$
Medium
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$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{a}}=\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}} \times 30=\frac{1.5}{1+1.5} \times 30$
$\frac{15 \times 30}{25}=3 \times 6$ or $V_{A}-V_{a}=18$ ...........$(1)$
also
$ \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{b}} =\frac{\mathrm{C}_{4}}{\mathrm{C}_{3}+\mathrm{C}_{4}} \times 30 $
$=\frac{0.5}{2.5+0.5} \times 30$
$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{b}}=\frac{0.5}{2.5+0.5} \times 10=5 \mathrm{V}$ .........$(2)$
equation $(1)$ and $ \mathrm{Eq}.$ $(2)$
$\mathrm{V}_{\mathrm{b}}-\mathrm{V}_{\mathrm{a}}=18-5=13 \mathrm{\,volt}$
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