Question

(Please show all work)

Each capacitor has a capacitance C = 12.0 ?F. (a) By combining capacitors in series and in parallel, find the

equivalent capacitance of this circuit. (b) If V = 10.0 volts, calculate Q, what would be the charge on each plate of

the equivalent capacitor?

C2 C1

Answer 1

3C 3C 8C Two capacitor with capacitances C and C) , set in parallel (the same Voltage across the ends) can be replaced by an equivalent, with Equivalent Capacitance given by Two capacitor with capacitances C1 and C2 , set in series (the same Current across the ends) can be replaced by an equivalent, with Equivalent Capacitance given by - or Ces 1 1 1 102 Ce 2 According to the text of the Problem " Each capacitor has a capacitance C - 12.0 uF". Then, all of the Capacitances have the same value C We will proceed step by step (see Figures above) applying the equations, to find the equivalent Capacitance, highlighted in green From original array (1° Figure to 2°), the two Capacitors, up to the left in serie, were replaced by one with Capacitance C/2 From 2° Figure to 3°, the two Capacitors, up to the right, in parallel, were replaced by one with Capacitance 3C/2 From 3° Figure to 4°, the two Capacitors, to the right, in serie, were replaced by one with Capacitance 3C/5 Finally, from 4° Figure to 5°, the two Capacitors, to the right and to the left of the Battery, in parallel, were replaced by the final one with Capacitance 8C/5 - C (a) Equivalent capacitance of the circuit. Cf-C , with the numerical value: C-12.0 , we have : 8 Cf )(12.0Jaf] G-19.211F] : Апswer to part(a) Of the Proble (b) From the definition of capacitance C - where Q is the Charge on the positive plate of the Capacitor ( Q in the negative plate) and V is the Voltage across the Capacitor ends. In this case: QCV .Substituting the numerical values, withV 10.0 [V] , we get Q-(19.2) 10.0 Q-192 [μ6] : Answer to part (b) of the Problem