Question 1[Maximum number: 2]Three capacitors are connected as shown in Fig. 4.1.Fig. 4.1Determine the total capacitance, in μF\mu \mathrm{F}μF, of the network of three capacitors.capacitance = μF\mu \mathrm{F}μFShow Answercombined capacitance of parallel capacitors =30(μ F)=30(\mu \mathrm{~F})=30(μ F) total capacitance =(1/45+1/30)−1=18μ F\begin{aligned} \text { total capacitance } =(1 / 45+1 / 30)^{-1} =18 \mu \mathrm{~F} \end{aligned} total capacitance =(1/45+1/30)−1=18μ FAdd to Test