Chapter 19 Lecture Notes


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1 Chapter 19 Lecture Notes Physics Strauss Formulas: R S = R 1 + R C P = C 1 + C /R P = 1/R 1 + 1/R /C S = 1/C 1 + 1/C q = q 0 [1e t/(rc) ] q = q 0 e t/(rc τ = RC 1. RESISTORS IN A CIRCUIT We will discuss electric circuits that have components connected in series and/or parallel. Series circuits are in connected in a straight line and have the same current through them. Parallel circuits have the same voltage drop across them. 1.1 Series Wiring Suppose two resistors are connected in series. The current going through each resistor is the same so V = IR 1 + IR 2 = I(R 1 + R 2 ) = IR s If circuit elements have no wire junctions between them, then they have the same current flowing through them and they are in series. To solve problems with simple circuits with resistors in series or parallel, first determine the equivalent resistance of the circuit, and then the total current flowing out of the battery. Problem: (a) What is the current flowing through a 3Ω, 6Ω, and 9Ω resistor in series when connected to a 12 V battery? (b) What is the voltage drop across each resistor? The equivalent resistance is 18 Ω which is greater than the resistance of any single resistor. We draw equivalent circuits with fewer resistors. 1.2 Parallel Wiring In a parallel circuit the voltage across each resistor is the same, so I = I 1 + I 2 = V/R 1 + V/R 2 =V(1/R 1 + 1/R 2 ) = V(1/R P ) In general, 1/R P = 1/R 1 + 1/R 2 + 1/R Problem: (a) What is the total current in a circuit if a 3Ω, 6Ω, and 9Ω resistor are connected in parallel to a 12 V battery? (b) What is the current in each resistor? 1
2 Note how the resistance of resistors in parallel is less than the resistance of any single resistor. 1.3 Parallel and Series Together (Equivalent Circuits) If I have resistors in parallel and in series, then I just take a step by step process to reduce this to an equivalent circuit. Problem: What is the equivalent resistance of this circuit? R 3 R 1 R 2 R 4 R 1 = 2 Ω R 3 = 5 Ω R 2 = 4 Ω R 4 = 10 Ω Demonstration/Problem: How many amps run through a tungsten wire when it is hooked up in series to a 100 watt light bulb and 600 watt heater that are in parallel? We can put many resistors together and often still reduce them to a single resistor. How can you rearrange the following diagram? All resistors having the same resistance. 2. KIRCHOFF S RULES Sometimes the circuit is too complicated to reduce to a single resistor, especially if there is more than one battery in the circuit. Then we use a set of rules, called Kirchoff s Rules, for determining the current through a circuit. 1. Junction Rule. The sum of the magnitudes of the currents directed into a junction equals the sum of the magnitudes of the currents directed out of the junction. (Conservation of charge) 2
3 2. Loop Rule. Around any closed loop, the sum of the potential drops equals the sum of the potential rises.. (Conservation of energy) How to Use Kirchoff s Rules: 1. Draw the circuit and draw currents with an arrow in every separate branch of the circuit. A branch is a section where the current does not change. 2. Apply the junction rule to enough junctions so that every current is used at least once. 3. Apply the loop rule to enough closed loops so that each current appears at least once. Remember the sign convention for the potential changes: Going across an battery is a plus sign if going from negative to positive and a minus sign if going from positive to negative. (Plus if going in normal direction of current.) Going across a resistor is a plus sign if going against the current and a minus sign if going with the current. (Minus if going in normal direction of current.) Problem: Find the current and power in the 3 Ω resistor. What is the current in the other two resistors? 6 Ω I 3 I 2 6 Ω 12 V I 1 3 Ω 3 V 3. TERMINAL VOLTAGE A real battery has some internal resistance. So if I were to draw a real battery it would consist of a perfect battery and a resistor. I can treat this just like two separate components when I use Kirchoff s laws or equivalent resistance. However, when I hook up to the battery I do not quite get the entire potential from the battery. Instead I get something slightly smaller. Problem: An ideal battery has a voltage of 9.0 volts and an internal resistance of 0.5 Ω. When it is drawing 0.3 amps, what is the terminal voltage? 3
4 4. CAPACITORS IN A CIRCUIT 4.1 Parallel When capacitors are in parallel, the total charge is the sum of the charge on each one, so... Q = Q 1 +Q 2 = C 1 V + C 2 V = V(C 1 + C 2 ) = C P V so C P = C 1 + C U = 1/2 C P V Series When capacitors are in series, the charge on each one is the same, so... V=Q/C 1 + Q/C 2 = Q(1/C 1 + 1/C 2 ) = QC s so 1/ C s = 1/C 1 + 1/C So, equivalent circuits for capacitors is done the same as for resistors, except that they combine differently. 5. RESISTORS AND CAPACITORS IN A CIRCUIT (RC CIRCUIT) If you attach a battery to a capacitor, the charge starts to build up on the capacitor. How fast it build up is given by the equation Q = Q 0 [1e t/(rc) ] where Q is the amount of charge on the capacitor at a time t, and Q 0 is the total amount of charge that the capacitor will hold. Since V = QC, this can also be written as V =V 0 [1e t/(rc) ] to determine the amount of voltage on the capacitor at any time. The value e is the natural logarithm. It is a transcendental number like π and it equals It can be found on most scientific calculators. y=e z. RC=τ is called the time constant. It is the time it takes for the capacitor to charge to 1/e of its total charge, or about 63.2%. The total charge on the capacitor is still given by Q 0 =CV. When discharging a capacitor, we disconnect the battery and let the charge flow of the capacitor, and we find Q = Q 0 e t/(rc) or in terms of voltage V = V 0 e t/(rc) and RC=τ is the time it takes for the capacitor to lose 63.2% of its charge. What if R is very small or 0, then e t/(rc) is zero and the charging and discharging happens instantaneously. 4
5 Problem : How long does it take for 50% of the maximum charge to be deposited on this circuit when the switch is closed. The resistor is 2 million ohms. and each capacitor is 10 nf. 6. ELECTRICAL SAFETY AND GROUNDING In a three prong outlet, the third prong makes a path of less resistance from the casing to the ground. Any current will flow through the third prong rather than through another body with more resistance. 5
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