0.75F
When capacitors are connected in series, the reciprocal of their total capacitance is the sum of the reciprocals of their individual capacitances. In other words, when, say, three capacitors, C₁, C₂ and C₃ are connected in series, their total capacitance, C, is given by;
[tex]\frac{1}{C}[/tex] = [tex]\frac{1}{C_1}[/tex] + [tex]\frac{1}{C_2}[/tex] + [tex]\frac{1}{C_3}[/tex] -------------------(i)
Now to solve the question,
Let;
C₁ = 1.0F
C₂ = 5.0F
C₃ = 8.0F
Substitute these values into equation (i) as follows;
[tex]\frac{1}{C}[/tex] = [tex]\frac{1}{1.0}[/tex] + [tex]\frac{1}{5.0}[/tex] + [tex]\frac{1}{8.0}[/tex]
Solve for C
[tex]\frac{1}{C}[/tex] = [tex]\frac{40.0 + 8.0 + 5.0}{40.0}[/tex]
[tex]\frac{1}{C}[/tex] = [tex]\frac{53.0}{40.0}[/tex]
C = [tex]\frac{40.0}{53.0}[/tex]
C = 0.75
Therefore, their total capacitance is 0.75F
