Answer:4.67 V
Explanation:
Given
[tex]C_1=5\ \muf[/tex]
[tex]C_2=7\ \muf[/tex]
Voltage [tex]V=8\ V[/tex]
For capacitance in series Net capacitance is given
[tex]C_{net}=\frac{C_1\cdot C_2}{C_1+C_2}[/tex]
[tex]C_{net}=\frac{5\times 7}{5+7}=\frac{35}{12} \muf[/tex]
[tex]q=C_{net}V[/tex]
[tex]q=\frac{35}{12}\times 8=\frac{70}{3} \muf[/tex]
As capacitors are in series so charge flow will be same in both capacitors
Voltage drop in [tex]5 \muf[/tex] capacitor
[tex]V=\frac{70}{3}\times \frac{1}{5}=\frac{14}{3} V\approx 4.67 V[/tex]
