Question

The capacitors have values C₁ = 2.5 uF and C₂ = 4.0 µF, C3 = 6.0 μF C₁ = 8.0 μF and the potential difference across the battery is 12.0 V. Assume that the capacitors are connected in series. Find the equivalent capacitance of the circuit and solve for the potential difference across each capacitors.

Answer 1

Answer

Equivalent capacitance is

$C=\frac{2.5*4*6*8}{4*6*8+2.5*6*8+8*2.5*4+2.5*4*6}\\=1.06\mu F$

Potential is given by

$V=\frac{q}{C}$