Physics 2: Capacitor Redux Kalamazoo Valley Community College John Stahl
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Physics 2: Capacitor Redux Kalamazoo Valley Community College John Stahl A voltage source provides charges which flow through the circuit. Circuits can be made of many different paths. The amount of charge flowing into the circuit depends on the network formed by the circuit elements. To solve for the amount of charge flowing into the circuit we need to take on the point of view of the voltage source, which cannot sense the different paths. The different paths can be combined and the complexity of the circuit reduced to a single component sensed by the source. In the following circuit reduce the network to solve for the single capacitor sensed by the source. The values of the circuit are listed in the table. C1 C2 C3 C4 C5 C6 C7 Vs 4.7mF 6.2mF 4.7mF 10mF 20mF 15mF 4.7mF 9v C1 Vs C4 C2 C3 C7 C5 C6 Before we can approach reducing the circuit we must deal with two capacitors that have no effect. C1 Vs C4 C2 C3 C7 C5 Open Circuit C6 Short Circuit An Open Circuit does not have a complete connection. One terminal dead ends, which takes the circuit element out of the reduction. This happens to C3. A Short Circuit occurs when a circuit element has the same wire connected across both ends. For charges to flow through the circuit element it needs to have a difference in voltage. When the same wire connects both ends the voltage is the same and no charges flow into it. We remove the elements that play no part in the circuit reduction. We start the circuit reduction by seeing the branches the charges flow through. C1 C4 Vs C2 C7 C5 There are two branches in the circuit. One branch contains C2 and the other branch contains C4 and C5. We want to reduce these branches so they only have a single element. In the branch on the right with C4 and C5 the same flow of charge passes into the capacitors. These are in series and we reduce them by fractional addition. 1 1 1 = + πΆππ πΆ4 πΆ5 1 1 1 = 10ππΉ + 20ππΉ πΆππ πΆππ = 6.67ππΉ Replace C4 and C5 with the Ceq and redraw the circuit. C1 Vs C2 6.67mF C7 With the two branches now having single elements each we see they span the same width, which makes them parallel. πΆππ = πΆ2 + 6.67ππΉ πΆππ = 6.2ππΉ + 6.67ππΉ πΆππ = 12.87ππΉ C1 12.87mF Vs C7 The final three capacitors are all in a continuous line, which makes them series. This will be the net capacitance seen by the voltage source. 1 1 1 1 = + + πΆππ πΆ1 12.87ππΉ πΆ7 1 1 1 1 = + + πΆππ 4.7ππΉ 12.87ππΉ 4.7ππΉ πΆππ = 1.99ππΉ
