For the system of capacitors shown in ?Fig. E24.16?, find the equivalent capacitance (a) between and c, and (b) between a and c.

Solution 16E If two capacitors with capacitances C and C ar1 connecte2 in parallel, then the equivalent capacitance is given by C eq = C +1C …..(2) If two capacitors with capacitances C and C are1connected2in series, then the C C equivalent capacitance is given by C eq = C + C2 …..(2) 1 2 (a) Between b and c, the capacitors with capacitances 9.0 pF and 11.0 pF are parallel. Hence, their equivalent capacitance would be C eq = 9.0 pF + 11.0 pF = 20.0 pF So, 20.0 pF is the equivalent capacitance between b and c. (b) Between the points a and c, the capacitor with the 15 pF capacitance will now be in series with the C eq(calculated in part a). This equivalent capacitance can be calculated as shown below. C eq1= 15×20pF = 8.6 pF 15+20 Therefore, 8.6 pF is the equivalent capacitance between a and c.