Question

Find the total capacitance of the combination of capacitors shown in Figure 2.5 μF 7. Find the total capacitance of the combination of capacitors shown in Figure 5.0 μ- 8.0 HF

please explain, thank you!

Answer 1

(6) When the capacitances are connected in series the equivalent capacitance is, 1 1 1 = = = +- сс, С, C= 99 - G+C₂ When the capacitances are connected in parallel the equivalent capacitance is, C=C+C The capacitors 0.3 uF and 10 uF are connected in series the equivalent capacitance is, C_(0.3 uF)(10 uF) 0.3 uF +10 uF = 0.291262 uF Now, the capacitance 0.291262 uF is connected in parallel to 2.5 uF, the equivalent capacitance of the circuit is, C' = 0.291262 uF +2.5 uF -2.8 uF Therefore, total capacitance of the circuit is 2.8 uF.

(7) Calculate the capacitance as follows: The capacitors 0.75 uF and 15 uF are connected in parallel the equivalent capacitance is, C=0.75 uF +15 4F = 15.75 uF The capacitors 1.5 uF and 15.75 uF are connected in series the equivalent capacitance is, _(1.5 uF)(15.75 uF) 1.5 °F+15.75 4F =1.3695652 uF C' The capacitors 5 uF and 3.5 UF are connected in series the equivalent capacitance is, _(5uF)(3.5 uF) 5 uF +3.5 4F = 2.0588 uF Now, the capacitances 2.0588 uF, 8 °F and 1.3695652 uF are connected in parallel the equivalent capacitance of the circuit is, Ceq = 2.0588 uF +8 uF +1.3695652 JF = 11.4 uF Therefore, total capacitance of the circuit is 11