Q.10. Two capacitors C1=3μF and C2=2μF are connected in series across d.c. source of 100 V. The ratio of the potential across C2 to C1 is
A. 2:3
B. 3:2
C. 6:5
D. 5:6
Answer :- B. 3:2
Explanation :-
\quad \mathrm{C}_1=3 \mu \mathrm{~F} \text { and } \mathrm{C}_2=2 \mu \mathrm{~F} \therefore \quad \mathrm{C}_{\text {series }}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}=\frac{6}{5} \mu \mathrm{~F} \mathrm{~A} = \mathrm{C}_{\text {series }} \times \mathrm{CV} \times \mathrm{V} .Q will be the same across both the capacitors as they are in series.
\therefore \quad Potential across capacitors,
\mathrm{V}_1=\frac{\mathrm{Q}}{\mathrm{C}_1}=\frac{120}{3}=40 \mathrm{~V} \mathrm{~V}_2=\frac{\mathrm{Q}}{\mathrm{C}_2}=\frac{120}{2}=60 \mathrm{~V} \therefore \quad \mathrm{~V}_2: \mathrm{V}_1=60: 40=3: 2