Answer:
[tex]C_{eq}=1.97\ \mu F[/tex]
Explanation:
Given that,
Capacitance 1, [tex]C_1=0.5\ \mu F[/tex]
Capacitance 2, [tex]C_2=11\ \mu F[/tex]
Capacitance 3, [tex]C_3=1.5\ \mu F[/tex]
C₁ and C₂ are connected in series. Their equivalent is given by :
[tex]\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}[/tex]
[tex]\dfrac{1}{C'}=\dfrac{1}{0.5}+\dfrac{1}{11}[/tex]
[tex]C'=0.47\ \mu F[/tex]
Now C' and C₃ are connected in parallel. So, the final equivalent capacitance is given by :
[tex]C_{eq}=C'+C_3[/tex]
[tex]C_{eq}=0.47+1.5[/tex]
[tex]C_{eq}=1.97\ \mu F[/tex]
So, the equivalent capacitance of the combination is 1.97 micro farad. Hence, this is the required solution.
