Explanation:
Capacitance, [tex]C_1=2.15\ F[/tex]
Capacitance, [tex]C_2=4.14\ F[/tex]
Capacitance, [tex]C_3=8.46\ F[/tex]
Capacitors C₁ and C₂ are connected in series. The equivalent capacitance is given by :
[tex]\dfrac{1}{C_s}=\dfrac{1}{C_1}+\dfrac{1}{C_2}[/tex]
[tex]\dfrac{1}{C_s}=\dfrac{1}{2.15}+\dfrac{1}{4.14}[/tex]
[tex]C_s=1.41\ F[/tex]
The combination is now connected with C₃ in parallel. Now the equivalent capacitance is given by :
[tex]C_p=C_s+C_3[/tex]
[tex]C_p=1.41+8.46[/tex]
[tex]C_p=9.87\ F[/tex]
So, the equivalent capacitance of the total combination is 9.87 F. Hence, this is the required solution.
