The capacitive network shown in the figure is assembled with initially uncharged capacitors. A potential difference, V_ab = +100V, is applied across the network. The switch S in the network is kept open. Assume that all the capacitances shown are accurate to two significant figures. What is potential difference V_cd across the open switch S? A) 0 V B) 40 V C) 50 V D) 60 V E) 70 V

Question

Quizplus · Accepted Answer

First, we can simplify the network by combining the capacitors in parallel and series. Note that C1 and C3 are in series with each other, so we can combine them into a single equivalent capacitor C1,3: 1/C1,3 = 1/C1 + 1/C3 1/C1,3 = 1/20 x 10^-6 + 1/60 x 10^-6 1/C1,3 = 1/15 x 10^-6 C1,3 = 15 x 10^-6 F Likewise, C4 and C5 are in series with each other, so we can combine them into a single equivalent capacitor C4,5: 1/C4,5 = 1/C4 + 1/C5 1/C4,5 = 1/15 x 10^-6 + 1/60 x 10^-6 1/C4,5 = 1/12 x 10^-6 C4,5 = 12 x 10^-6 F Now we have two capacitors in parallel: C1,3 and C2, and C4,5 and C6. We can combine these into single equivalent capacitors: Ceq1 = C1,3 + C2 Ceq1 = 15 x 10^-6 + 20 x 10^-6 Ceq1 = 35 x 10^-6 F Ceq2 = C4,5 + C6 Ceq2 = 12 x 10^-6 + 30 x 10^-6 Ceq2 = 42 x 10^-6 F We can now redraw the network with these equivalent capacitors: _______ | | V_{---Ceq1---+ +---Ceq2---S---V} |_______| Before the switch is closed, the potential difference across each capacitor is 0 V. When the switch is closed, the capacitors will start to charge, but the charge is distributed in such a way that the potential difference across each branch of the network (i.e. each capacitor) remains constant. Let Q be the charge on each capacitor in the network after it has been charged. Then, by conservation of charge, we have: Q = Ceq1 x V1 = Ceq2 x V2 where V1 and V2 are the potential differences across Ceq1 and Ceq2, respectively. We know that V1 + V2 = 100 V (from the potential difference across the battery), so we can express V1 in terms of V2: V1 = 100 V - V2 Substituting into the equation above, we get: Q = Ceq1 x (100 V - V2) = Ceq2 x V2 Solving for V2, we get: V2 = 100 V x (Ceq1 / (Ceq1 + Ceq2)) V2 = 100 V x (35 x 10^-6 F / (35 x 10^-6 F + 42 x 10^-6 F)) V2 = 40 V Therefore, the potential difference across the open switch S is equal to the potential difference across Ceq2, which is equal to V2, or 40 V. Hence, the answer is (B) 40 V.

The Capacitive Network Shown in the Figure Is Assembled with Initially