Physics - Capacitor Question with Solution | TestHub

given: $C_S=3Mf$ $\begin{array}{rl}& C_P=16Mf\\ & \text{as }{C}_S=\left[\left(C_1{C}_2\right)/\left(C_1+C_2\right)\right]=3\\ & C_P=C_1+C_2=16(2)\\ & \therefore \text{from (1), }\left[\left(C_1{C}_2\right)/(16)\right]=3\\ & C_1{C}_2=48\\ & \text{from (2), }{C}_2=16-C_1\\ & C_2=16-\left(48/C_2\right)\\ & \therefore C_2+\left(48/C_2\right)=16\\ & \therefore C_2^2+48-16C_2=0\\ & \therefore C_2=12\mu F,C_2=4\mu F\end{array}$ i.e. capacitance are $4\mu \mathrm{F}\&12\mu \mathrm{F}$