Question 5.6: Capacitors are connected in a circuit as shown in Fig. 5.11.......

Capacitors 3 and 5 μF are connected in series. Here, the total capacitance is,

C_{1}={\frac{3\times5}{3+5}}=1.88\,\mu\mathrm{F}         (5.80)

The capacitors 1.88 and 8 μF are connected in parallel, and the total capacitance is,

C_{2}=8+1.88=9.88\,\mathrm{\mu F}         (5.81)

Again, 9.88 and 2 μF are connected in series and the equivalent circuit capacitance is,

C_{\mathrm{eq}}={\frac{9.88\times2}{9.88+2}}=1.66\,\mu\mathrm{F}         (5.82)