# Circuits. 1. The Schematic

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1 + ircuits 1. The Schematic 2. Power in circuits 3. The Battery 1. eal Battery vs. Ideal Battery 4. Basic ircuit nalysis 1. oltage Drop 2. Kirchoff s Junction Law 3. Series & Parallel 5. Measurement Tools 1. The mmeter 2. oltmeter 6. ircuits 1. Discharging a capcitor 2. harging a capacitor 3. circuit 7. Grounding Here is the circuit diagram for an electronic keyboard. It's essentially a map that shows the layout of all the circuit elements. Just like when there is an accident or construction event on a city street, consulting the map can help diagnose and repair the issue. 1. The Schematic ircuit Diagram is a schematic pf Ω 200 Ω 200 Ω 2 pf By agreeing on a 'shorthand' or system of symbols and rules, we can make circuits much more approachable and understandable. updated on J. Hedberg 2018 Page 1

2 + Battery Lamp/Bulb esistor Junction No apacitor Switch onnection Here are the standard circuit elements for that we'll work with. 2. Power in circuits We start with the potential energy of a charge: In differential form, this would be But current is defined by: I = dq/dt, so we write: Power has always been defined as the rate of energy change: du/dt Thus, we can say for power associated with transferring the energy (of chemical in a battery for eg.) into electrical energy is s usual, the unit of power is the Watt U = q du = dq du = Idt P = I Power in circuits In this form, P = I, we're really asking how fast a battery or solar cell or wind turbine is transforming one form of energy into electrical potential energy. Using Ohm's Law, = I, we can rewrite this equation as: or P = These expressions can tell us about the transfer of energy in a resistive device, like a lightbulb. The electrical potential energy of the charges passing through a resistive device will get transformed into another form, which in most cases is thermal energy. I 2 P = 2 updated on J. Hedberg 2018 Page 2

3 Example Problem #1: The voltage difference in an electrical outlet in the home is 120. What should the resistance of a 100 W light bulb be? How long and thick should the tungsten filament be to provide such a resistance? Quick Question 1 The insulated wiring in a house can safely carry a maximum current of 18. The electrical outlets in the house provide a voltage of 120. space heater when plugged into the outlet operates at an average power of 1500 W. How many space heaters can safely be plugged into a single electrical outlet and turned on for an extended period of time? The Battery Before we can do anything, we need to separate charges. If a device can maintain this separation, then we'll say it can generate an EMF, an electromotive force, E We'll start by talking about batteries as the main generator of EMF, but later we'll see several other, more common and more complicated sources. Essentially, all the EMF sources are similar in that they can keep two pieces of metal at constant, but different, potentials for some time. EMF E = dw dq The definition of EMF, E, is the work per unit charge that the device does while moving charge from the low potential terminal to the high potential terminal. The SI unit for this would then be Joules/oulomb. (We called this for olt before, so let's stay with that) updated on J. Hedberg 2018 Page 3

4 + eal Battery vs. Ideal Battery real battery r In an ideal battery, there will be no internal resistance. In a real battery, there will be a resistance due to the metal of the terminals and other parts. These are built into the battery and are not removeable. ny real EMF device will have an internal resistance. You can assume batteries are ideal, unless otherwise indicated. 4. Basic ircuit nalysis Two tools: current and voltage oltage Drop resistor This is an important concept: When analyzing a circuit, we can talk about the voltage drops from point to point. That is, what s the electric potential at various locations in the circuit. Since we ve said the wires are ideal conductors, no voltage drops will occur along the wire. In this simple circuit, the voltage drop happens in the resistor. x Quick Question B resistor ank the electric potential at the five points in the figure 1. > B > > D > E 2. B > D = > > E 3. B = > > D = E 4. B = E > = D > 5. we can t say since we don t know the value of Ω 1.5 D E updated on J. Hedberg 2018 Page 4

5 Kirchoff s Junction Law This is consequence of the conservation of charge. We can use it to figure out the current in any branch of the circuits. I in I in = I out I in = I 1 + I 2 + I 3 I 1 I 3 I 2 Kirchoff s Loop Law The sum of voltage rises/drops around a loop is zero. Meaning, if we add all the potential differences around a loop, the sum should equal zero. For every uphill there is a downhill. [emember gravity? It was a lot like this too.] Battery Series & Parallel lamps in series lamps in parallel If two components are connected in series, then there are no junctions between them. If two components are connected by a junction at their ends, then they are connected in parallel. These two methods of forming circuits have very different implications for how charges will move and the performance of the circuit. updated on J. Hedberg 2018 Page 5

6 resistors in series or Here we have two resistors, and, connected to a battery in series. Starting at the negative terminal of the battery, we can begin adding up all the changes in potential along the circuit. [Kirchoff s loop law] Δ = E = +Δ 1 + Δ 2 = 0 Since Δ 1 = I 1 and Δ 1 = I 2, we can use the K.L.L. to find the current in the circuit: E = Δ 1 Δ 2 = I 1 + I 2 I = E resistors in series resistors in series + equivalent resistance This equation for the current suggests we can just add the series resistors together, to create an equivalent resistance for the circuit. This is true. The eq is the sum of all resistors in series. E E I = I = eq eq = eq n = i=1 n updated on J. Hedberg 2018 Page 6

7 Parallel omponents resistors in parallel Now we have some junctions to deal with. I in = I out I in I 1 I 2 equal Below is the same circuit, in a different style. First we can use Ohm s law for each resistor to see the total current coming from the battery Δ I bat = 1 Δ I bat E E 1 1 = + = E ( + ) Parallel omponents resistors in parallel The current, I, coming from the battery 1 1 I bat = E ( + ) 1 2 gain, we can think of an equivalent resistancewhich can be used to replace the two individual resistances: I in equal I 1 I 2 eq Δ E 1 1 = = = ( + ) I I bat eq = ( ) eq n = i=1 1 n N 1 1 updated on J. Hedberg 2018 Page 7

8 Quick Question 3 Two light bulbs (not necessarily the same ) can be connected in series or parallel. In each case, what quantity is the same: hoice Parallel Series Potential Difference across each bulb Potential Difference across each bulb B urrent through each bulb Potential Difference across each bulb urrent through each bulb urrent through each bulb D Potential Difference across each bulb urrent through each bulb Brightness in the bulbs The voltage source and bulb resistances are all the same, which set will have the brightest bulbs? (Power determines brightness) Quick Question 4 B When the switch closes, what happens to the brightness in the bulbs: 1. The brightness of the bulbs is not affected. 2. Bulb becomes brighter, bulb B goes out. 3. Bulb B becomes brighter, bulb goes out. 4. Both bulbs become brighter. Quick Question 5 B Describe the brightness in the bulbs, if all three have the same resistance: 1. Bulbs B and are equally bright, but bulb is less bright. 2. Bulbs B and are equally bright, but less bright than bulb. 3. ll three bulbs are equally bright. 4. Bulbs and B are equally bright, but bulb is less bright. 5. Only bulb is illuminated. updated on J. Hedberg 2018 Page 8

9 5. Measurement Tools ammeter voltmeter points on a circuit. Two more symbols: The ammeter is a tool that measures current through a point on a circuit. The voltmeter measures the potential difference between two Understanding how to use these tools requires understanding what they are trying to measure. The mmeter What s the current here? The ammeter is placed in series. This means any resistance it has will contribute to the eq of the circuit. Thus: mmeters are built to have near zero resistance. The ideal ammeter would therefore have = 0 Ω. ammeter Put an ammeter there to find out. nd thus, by measuring I in this simple 2 resistor circuit, we would be able to find the total eq for this circuit. eq E = = I I ould we use the ammeter to find 1 and 2 (without altering the circuit)? I I OM To install the ammeter in series, simply add it to the circuit and ensure that one wire goes to the I input jack, and the other goes to the OM jack. This setup measures the current between the two resistors. The measured I will depend on the battery and both resistors in the circuit. updated on J. Hedberg 2018 Page 9

10 + oltmeter voltmeter connected across 2 will tell us the potential difference between the two sides of the resistor. 1 2 What s between these points? Battery onnect a voltmeter across. dding a voltmeter to the circuit is like adding another component in parallel. So, its internal resistance has to be very large (infinity in the ideal case) to prevent it from changing the of the circuit. eq = ( ) eq 1 2 v-meter 1.5 I I OM voltmeter gain, we must now change the setting, and rearrange our connections between the voltmeter and the circuit. Now, the two leads go into the jack and the OM jack. Now, we can measure the voltage drop across. 2 updated on J. Hedberg 2018 Page 10

11 Example Problem #2: What would the ammeter display in these situations Ω Ω 20 Ω 20 Ω Ω 20 Ω 3 6. ircuits Our original capacitor had two plates, separated by a vacuum. -plate +plate -plate +plate 0 + vacuum 0 + insulator real capacitor has an insulator (dielectric) stuffed in between the two plates. updated on J. Hedberg 2018 Page 11

12 apacitors in Parallel Here we have 3 capacitors arranged in parallel, 1, 2, and 3 all connected to one battery with voltage. If they each have the same capacitance, then the total charge stored will be 3 times greater than if there were only one apacitor. If they are different, then to calculate an equivalent capacitance, we just need to add them all up. So for, N capacitors, we can say: eq = N = N i=1 i apacitors in Series 1 2 If however, we have 2 or more capacitors in series, then things are a bit different. We can ask what the eq would be for this configuration. 1 eq = Δ Q The charge Q will be the same on each capacitor, but each individual one might have a different Δ, depending on its individual capacitance. 1 Δ = 1 Δ = + eq Q Q eq eq N 1 = i=1 i updated on J. Hedberg 2018 Page 12

13 Discharging a capcitor B open closed closed Here are four snapshots of what happens when a capacitor discharges.. The capacitor is fully charged, but the switch is open. No current is present in the circuit. B. The switch is closed and charge flows.. fter some time, the amount of charge imbalance between the capacitor has diminished. Thus its Δ is smaller, and thus so is the current through the circuit. D. The Δ is now almost zero, so nearly no current is present. I D closed B D t B D t harging a capacitor open Let s go the other way, and see how a battery will charge a capacitor. B D closed closed closed 0 I. gain, we start with an open switch. No current/no Δ B. losing the switch allows charges to flow from the battery to the capacitor. The current is large at this point, but Δ is still small. fter some time, the current has decreased, but Δ is increasing D. Finally, the capacitor is essentially fully charged, and no current is present in the circuit, and Δ = the Δ of the battery. B D t 0 B D t updated on J. Hedberg 2018 Page 13

14 circuit open closed Now, we include both a resistor and a capacitor in the same circuit. If we start at the battery (low side) and count all the potential differences, we get: But, I = dq/dt, so: Q E = E I = 0 dq E = + dt Q which is a differential equations. We'll need to solve it for Q. Q = E (1 e t/ ) harging an circuit Does this make sense? Q = E (1 e t/ ) Let's take the time derivative of this formula to find the current as a function of time. dq E I = = ( ) dt e t/ I closed t = 0 t updated on J. Hedberg 2018 Page 14

15 Time onstant x = +1 I The function e t/1 will be equal to 1/e when t = 1. I =.367 I 0 I =.135 I 0 t = 0 t In the case of our charging capacitor, the equation is e t/. Thus, we'll establish a "time constant" given by τ, which will equal. nd thus, after 1 time constant, the value of the current passing through this circuit will be 37% of what it was at t = 0. Thus, τ = is a way of describing the nature of the circuit. Discharging an circuit battery connected discharge closed Just like before, we can look at a path around a discharging circuit and add up all the voltage drops (or rises): It will be similar to before, except now the E (emf) will be 0: dq Q I + Q/ = 0 + = 0 dt updated on J. Hedberg 2018 Page 15

16 Discharging an circuit dq dt Q = This is another differential equation, the solution (Q as a function of t) of which is: where Q 0 = 0 Q = Q 0 e t/ is the initial charge on the capacitor Taking the time derivative of this, we can obtain the current as a function of time: dq Q I = = ( 0 ) dt e t/ Discharging an circuit Does this equation make sense? Q I = ( 0 ) e t/ updated on J. Hedberg 2018 Page 16

17 7. Grounding circuit 1 circuit 2 are these two potentials the same? circuit 1 circuit 2 Now, both zero points are defined by potential of the earth. Ground Loops updated on J. Hedberg 2018 Page 17

18 Example Problem #3: Find the potential difference across, and the current through, each resistor Ω 4 Ω 8Ω Find the potential difference across, and the current through, each resistor Ω 4 Ω 8Ω updated on J. Hedberg 2018 Page 18

19 Example Problem #4: What is the equivalent resistance of this group of resistors 10 Ω 25 Ω 90 Ω 45 Ω What is the equivalent resistance of this group of resistors 10 Ω 25 Ω 90 Ω 45 Ω updated on J. Hedberg 2018 Page 19

20 Example Problem #5: Find the potential difference across, and the current through, each resistor. 600 Ω Ω 400 Ω 560 Ω Find the potential difference across, and the current through, each resistor. 600 Ω Ω 400 Ω 560 Ω Example Problem #6: + The batteries in the figure are identical. If the currents passing through each resistor are equal, what is the resistance of the second resistor? +? updated on J. Hedberg 2018 Page 20

21 Example Problem #7: 100 Ω What voltage will make 1 out = 10 in Example Problem #8: What is the potential drop across? ( = = 3 = 500 Ω.) updated on J. Hedberg 2018 Page 21

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