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Published byMarc Burbridge Modified about 1 year ago

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Goal: To understand how Capacitors work in circuits. Objectives: 1)To learn how Capacitors work in parallel 2)To learn how Capacitors work in series 3)To learn how Capacitors work in Combinations of series and parallel 4)To learn about RC Circuits

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Capacitors in parallel Capacitors are like dams for charge. Imagine a stream split, and you had two dams side by side. These dams would be parallel and independent (somewhat) of one another. The water you could store would just be the sum of what each dam stored. Capacitors in parallel are just like that.

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Effective Capacitance The effective capacitance for the circuit is the sum of each capacitance. So, C = C1 + C2 + … + Clast Lets try one. You have 3 capacitors in parallel. Their Capacitances are 1F, 10F, and 5F. What is the effective capacitance of the circuit?

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Capacitors in series This one is a bit trickier. This one is like a series of dams. Each one affects the next. In this case the inverses add. So, the effective capacitance is: 1/C = 1/C1 + 1/C2 + … + 1/Clast

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Sample problem Lets do 2 in series Capacitors. The capacitors have capacitances of 5 F and 3 F. What is the effective capacitance?

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Sample problem Lets do 2 in series Capacitors. The capacitors have capacitances of 5 F and 3 F. What is the effective capacitance? 1/C = 1/C1 + 1/C2 = 1/5 + 1/3 Note there is a trick you can use here! (shown on board) Note the capacitance actually goes down! This is because we are less efficient in storing charge.

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Hybrid circuits Often times you will get a hybrid circuit. This is a circuit that has components that are in series and in parallel. You have to break circuits like this into parts and solve the parts. To do this try to find something you can isolate, then solve for that part, and replace. Lets try one together.

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Hybrid circuits For the circuit shown C1 = 5F, C2 = 3F, and C3 = 10F. What can we isolate first? Are they in series or in parallel?

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Hybrid circuits For the circuit shown C1 = 5F, C2 = 3F, and C3 = 10F. What can we isolate first? We isolate C1 and C2. They are in parallel so we add them. So, C12 = C1 + C2 = 8F.

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Hybrid circuits For the circuit shown C1 = 5F, C2 = 3F, and C3 = 10F. C12 = 8F Now we have a circuit with 2 capacitors effectively, C3 and C12. Are they in series or in parallel?

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Hybrid circuits For the circuit shown C1 = 5F, C2 = 3F, and C3 = 10F. C12 = 8F Now we have a circuit with 2 capacitors effectively, C3 and C12. Since they are in series we add by inverse. Using the trick we got before to do this faster we get that C = (8 * 10) / (8 + 10) F Or C = 4.4 F

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Lets try 1 more. This one is similar to the one on the homework, but not exact: (drawn on board) C1 = 5 F, C2 = 1 F C3 = 8 F, C4 = 4F What is the effective capacitance (break into parts!)?

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Lets try 1 more. C1 = 5 F, C2 = 1 F C3 = 8 F, C4 = 4F What is the effective capacitance (break into parts!)? We can combine the two parts that are each in parallel – C1/C2, and C3/C4. C12 = C1 + C2 = 6F C34 = C3 + C4 = 12F

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Lets try 1 more. What is the effective capacitance (break into parts!)? C12 = C1 + C2 = 6F C34 = C3 + C4 = 12F We now have 2 capacitors in series, C12 and C34. So, C = (C12 * C34) / (C12 + C34) C = 72 / 18 F= 4F

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RC Circuits If you have a circuit with a resistor and a capacitor you have a RC circuit. Resistors are objects that resist the flow of charge. The rate of flow of charge is called current RC circuits allow you to control the current. Therefore they are used in timing devices.

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Charging a capacitor in a RC circuit In a RC circuit the capacitor charges exponentially. That is, over a time constant called the “time constant” a fraction of the max allowed charge is added. The amount added after the first time constant is: Q = Qmax * (1 – e -1 ) Here Qmax is the max charge (or Qmax = CV), and e is the natural log (e 1 = 2.71)

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Charging a capacitor in a RC circuit In a RC circuit the capacitor charges exponentially. The time constant is: τ = RC And Q = Qmax * (1 – e -t/τ ) Also note that the ln(e x ) = x Also, Vcapacitor = Vbattery * (1 – e -t/τ ) And the current is: I(t) = (Vbattery / R) * e -t/τ

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Try one: Similar to HW2: You have a resistor of 0.9 Ohms and a 5.2 F capacitor connected to a 28 V battery. Find: A) The time constant B) The maximum charge on the capacitor C) Use your answer from A and B to find the amount of charge on the capacitor after 12 sec.

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Try one: Similar to HW2: You have a resistor of 0.9 Ohms and a 5.2 F capacitor connected to a 28 V battery. Find: A) The time constant τ = RC = 0.9 * 5.2 = 4.68 s B) The maximum charge on the capacitor Q = CV = 5.2 * 28 = C C) Use your answer from A and B to find the amount of charge on the capacitor after 12 sec. Q = Qmax * (1 – e -t/τ ) = C * ( 1 - e -12/4.68 ) Q = 134 C

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One more: Resistor is 2.6 Ohms Capacitor is 2.3 F Battery is 10 V How long will it take to charge to 90% of maximum? Hint: what is the value of (1 – e -t/τ ) here? 2 nd Hint if you get that A = e b Then B = ln(a)

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One more: Resistor is 2.6 Ohms Capacitor is 2.3 F Battery is 10 V How long will it take to charge to 90% of maximum? (1 – e -t/τ ) = 0.9 (thus 90%) So, e -t/τ = 0.1, and –t/ τ = ln(0.1) = -2.3 So, t = 2.3 * τ τ = RC = 2.6 * 2.3 = 5.98 So, t = 2.3 * 5.98 = 13.8 s

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conclusion We learned how to find effective capacitances for captors in parallel (sum) and in series (sum of inverses). We learned how to break hybrid circuits into parts. We learned how to calculate the time constant (τ = RC) and how this applied to the charging of a capacitor in a RC circuit.

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