The equivalent capacitance between A and B is (in $\mu \mathrm{F}$ )\

Question

Professor PC Thomas Classes · Answer

A starting from the right end of the network three $3 \mu \mathrm{~F}$ capacitors are connected in series. The equivalent capacitance of these three capacitors is\ $\frac{1}{\mathrm{C}_{\mathrm{s}}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3} \Rightarrow \mathrm{C}_{\mathrm{s}}=1 \mu \mathrm{~F}$\ $\mathrm{C}_{\mathrm{s}}$ and $2 \mu \mathrm{~F}$ are in parallel. The equivalent capacitance of these two capacitors is\ $\mathrm{C}_{\mathrm{P}}=\mathrm{C}_{\mathrm{S}}+2 \mu \mathrm{~F}=1 \mu \mathrm{~F}+2 \mu \mathrm{~F}=3 \mu \mathrm{~F}$\ Proceeding in this way, finally three $3 \mu \mathrm{~F}$ are in series. Therefore, the equivalent capacitance between A and B is\ $\frac{1}{\mathrm{C}_{\mathrm{AB}}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3} \Rightarrow \mathrm{C}_{\mathrm{AB}}$\