Circuit Diagrams Circuit diagrams are ways in which we can draw a circuit with all of its components The components that we will see are: Battery: Applies a voltage across the circuit to create current Resistor: Impedes current Capacitor: Stores Charge Ground Switch: Opens and closes the circuit Circuit Diagrams To read a circuit diagram, start with the battery (the longer side indicates the positive end of the battery) and follow the straight lines which represent basic wires The lines will lead into other circuit components Remember that we must always complete the circuit, so the wires must lead back to the battery in the end We see that a wire comes from the battery, passes through a resistor, then a capacitor, then leads back into the battery to complete the loop This is an example of a simple RC circuit Wire Wire You will construct these yourself in Lab 5 Resistor Capacitor Circuits – Resistors in Series There are different ways in which we can put resistors together in a circuit to either increase or decrease the total resistance When two circuit components are put together is series, this means that one wire leads directly into the next circuit component A circuit with 3 resistors in series All resistors create a voltage drop, and that the total voltage of the battery is equal to the voltage drop across all of the resistors A property of circuits elements in series is that the current is the same through the whole circuit Circuits – Resistors in Series If we utilize these two properties, then we can set the total voltage of the battery equal to the sum of voltage drops in each resistor ππ‘ππ‘ = π1 + π2 + π3 ⇒ π 3 πΌπ π‘ππ‘ = πΌπ 1 + πΌπ 2 + πΌπ 3 πΌ πΌ Since the current is the same through the whole circuit, we can divide it out π π‘ππ‘ = π 1 + π 2 + π 3 This means that the total resistance in the circuit is equal to the sum of all the resistance in the resistors π π‘ππ‘ = π 1 + π 2 + π 3 + β― π 1 π 2 Circuits – Resistors in Parallel A second way that circuit elements can be set up is in parallel. You may see this drawn a few ways that all mean the same thing 12 π π΄ 12 π π΄ π΅ 12 π π΅ 12 π The key feature in parallel circuits is that there is a junction where current splits into different branches of the circuit A property of parallel circuits is that the voltage in each branch is the same and equal to the voltage of the battery Circuits – Resistors in Parallel It is important to label the current in each branch properly because current splits in each branch πΌ2 πΌ1 πΌπ‘ππ‘ πΌπ‘ππ‘ It can be easiest to define the total current as the current leaving the battery The current then spits into branch 1 and branch 2. We can label these currents as πΌ1 and πΌ2 What is the do we label the current here? The current at the end of the circuit should equal the current at the beginning of the circuit Circuits – Resistors in Parallel Electrical currents behave very similarly to water in a river. If water is moving down a river and the river splits, then all the water must be split into the two streams πππ‘ππ1 πππ‘ππ2 The total amount of water entering the split is equal to the amount of water in both streams π΄ πΌπ‘ππ‘ πππ‘πππ‘ππ‘ = πππ‘ππ1 + πππ‘ππ2 πππ‘πππ‘ππ‘ Electric current works the same way. The total current, πΌπ‘ππ‘ comes from the battery, and splits into each branch πΌπ‘ππ‘ = πΌ1 + πΌ2 πΌ1 πΌ2 π΅ Circuits – Resistors in Parallel If we apply Ohm’s Law (again), πΌ = π/π , we can transform this equation πΌπ‘ππ‘ = πΌ1 + πΌ2 π π π = + π π‘ππ‘ π 1 π 2 π΄ πΌπ‘ππ‘ Remember that the voltage in each branch of a parallel circuit is the same so the voltages will cancel out 1 1 1 = + π π‘ππ‘ π 1 π 2 1 1 1 = + +β― π π‘ππ‘ π 1 π 2 πΌ1 πΌ2 π΅ Example 1– Series and Parallel Examine the following circuit and find the total resistance as well as the total current drawn from the battery π 2 = 500 Ω π 1 = 400 Ω Strategy: We want to try to combine the resistors until it becomes one giant resistor 1) Start by combining parallel components π 3 = 700 Ω 12 π 1 1 1 = + π β₯ 500 Ω 700 Ω 1 1,200 = = 0.00343 Ω−1 π β₯ 350,000 Ω π β₯ = 292 Ω Example 1 – Series and Parallel Examine the following circuit and find the total resistance as well as the total current drawn from the battery π 1 = 400 Ω π β₯ = 292 Ω Strategy: We want to try to combine the resistors until it becomes one giant resistor 1) Start by combining parallel components When we combine parallel resistors, they look like just one resistor that is now in series with π 1 12 π 2) Now combine the resistors that are in series π π‘ππ‘ = π 1 + π β₯ π π‘ππ‘ = 400 Ω + 292 Ω π π‘ππ‘ = 692 Ω Example 1 – Series and Parallel Examine the following circuit and find the total resistance as well as the total current drawn from the battery Now we want the total current from this battery. This will utilize Ohm’s Law, and use the total resistance to find the total current π π‘ππ‘ = 692 Ω π = πΌπ‘ππ‘ π π‘ππ‘ πΌπ‘ππ‘ πΌπ‘ππ‘ 12 π Total current can be found in the wire on either side of the battery πΌπ‘ππ‘ = π/π π‘ππ‘ πΌπ‘ππ‘ = 12 π 692 Ω πΌπ‘ππ‘ = 0.0173 π΄ Example 2 – Series and Parallel Examine the following circuit and find the total resistance as well as the current through π 2 π 5 = 10.0 Ω π 3 = 8.0 Ω π 2 = 6.0 Ω 1 1 1 = + π 34 8.0 Ω 4.0 Ω 1 = 0.375 Ω−1 π 34 π 4 = 4.0 Ω π 1 = 5.0 Ω Let’s start to combine resistors, staring with the simplest parallel circuit π = 9.0 π π 34 = 2.7 Ω Example 2 – Series and Parallel Examine the following circuit and find the total resistance as well as the current through π 2 π 5 = 10.0 Ω Next let’s combine the π 2 and π 34 which are in series with each other π 234 = 6.0 Ω + 2.7 Ω π 2 = 6.0 Ω π 1 = 5.0 Ω π 34 = 2.7 Ω π = 9.0 π π 234 = 8.7 Ω Example 2 – Series and Parallel Examine the following circuit and find the total resistance as well as the current through π 2 π 5 = 10.0 Ω Now π 5 and π 234 are in parallel with each other 1 = 1 1 + 10.0 Ω 8.7 Ω = 18.7 87 Ω π 2345 = 87 Ω 18.7 π 2345 π 234 = 8.7 Ω 1 π 2345 π 1 = 5.0 Ω π = 9.0 π π 2345 = 4.7 Ω Example 2 – Series and Parallel Examine the following circuit and find the total resistance as well as the current through π 2 The total resistance is then π π‘ππ‘ = π 1 + π 2345 = 5.0 Ω + 4.7 Ω π 2345 = 4.7 Ω π π‘ππ‘ = 9.7 Ω The total current is then πΌπ‘ππ‘ π 1 = 5.0 Ω π = π π‘ππ‘ πΌπ‘ππ‘ = 0.93 π΄ π = 9.0 π How do we find the current in π 2 ? Let’s go back a step and investigate Example 2 – Series and Parallel Examine the following circuit and find the total resistance as well as the current through π 2 π 5 = 10.0 Ω πβ₯ πΌ5 The voltage across the parallel part will be the same in both branches which is equal to the voltage of the battery minus the voltage drop across π 1 πβ₯ = 9.0 π − πΌπ‘ππ‘ π 1 πβ₯ = 9.0 π − 0.93 π΄ 5.0 Ω πΌ2 π 1 = 5.0 Ω π 234 = 8.7 Ω π = 9.0 π πβ₯ = 4.35 π This is the voltage left that must go through the parallel portion πβ₯ = πΌ2 π 234 πΌ2 = 4.35 π πΌ2 = 8.7 Ω πΌ2 = 0.5 π΄ π π 2 Kirchhoff's Rules The past two examples have been examples of simple circuits. Sometimes circuits can get too complicated, and we need a set of rules that can apply to any type of circuit Lucky for us, in 1845, Gustav Kirchhoff developed a set of rules to find the voltage, resistance, and current in any branch of a circuit 1. Loop Rule / Voltage Rule “The sum of the changes in potential around any closed loop of a circuit must be zero” πππππ = 0 1. Junction Rule / Current Rule “At any junction point, the sum of all currents entering the junction must equal the sum of all currents leaving the junction” πΌππ = πΌππ’π‘ Gustav Kirchhoff 1824 - 1887 Example - Kirchhoff's Rules Determine the magnitudes and directions of the currents in each resistor using Kirchhoff ’s rules π = 9.0 π πΌ1 π 1 = 25.0 Ω It is helpful to first label the current in each branch as well as direction. The direction can be guessed! If you happened to pick the wrong direction, your math will give you a negative value π 2 = 68.0 Ω Our unknowns are the currents in each branch, πΌ1 , πΌ2 , and πΌ3 so we will set up Kirchhoff's rules to solve for them πΌ2 π = 12.0 π π 3 = 35.0 Ω πΌ3 Our end goal is to set up a group of equations that can be substituted into each other Example - Kirchhoff's Rules Determine the magnitudes and directions of the currents in each resistor using Kirchhoff ’s rules πΌ1 Junction Rule: One equation for each junction π 1 = 25.0 Ω π΄: πΌ2 = πΌ1 + πΌ3 B: πΌ1 +πΌ3 = πΌ2 π1 = 9.0 π Loop Rule: We need to create loops and follow the voltages π 2 = 68.0 Ω π΄ 1: In Loop 1, we start at the battery and move across π 1 then π 2 in the direction of the current, resulting in a Voltage Drop π΅ πΌ2 π2 = 12.0 π π 3 = 35.0 Ω πΌ3 1: π1 − πΌ1 π 1 − πΌ2 π 2 = 0 Example - Kirchhoff's Rules Determine the magnitudes and directions of the currents in each resistor using Kirchhoff ’s rules πΌ1 Junction Rule: One equation for each junction π 1 = 25.0 Ω π΄: πΌ2 = πΌ1 + πΌ3 B: πΌ1 +πΌ3 = πΌ2 π1 = 9.0 π Loop Rule: We need to create loops and follow the voltages π 2 = 68.0 Ω π΄ 1: In Loop 1, we start at the battery and move across π 1 then π 2 in the direction of the current, resulting in a Voltage Drop π΅ πΌ2 π2 = 12.0 π π 3 = 35.0 Ω πΌ3 1: π1 − πΌ1 π 1 − πΌ2 π 2 = 0 2: In Loop 2, we start at the battery and move across π 2 then π 3 against the direction of the current, resulting in a Voltage Gain 2: −π2 + πΌ3 π 3 + πΌ2 π 2 = 0 Example - Kirchhoff's Rules Determine the magnitudes and directions of the currents in each resistor using Kirchhoff ’s rules π΄: πΌ2 = πΌ1 + πΌ3 πΌ1 π 1 = 25.0 Ω B: πΌ1 +πΌ3 = πΌ2 1: π1 − πΌ1 π 1 − πΌ2 π 2 = 0 2: −π2 + πΌ3 π 3 + πΌ2 π 2 = 0 π1 = 9.0 π There is a last loop that goes around the outside of the whole circuit π 2 = 68.0 Ω π΄ π΅ πΌ2 π2 = 12.0 π 3: π1 − πΌ1 π 1 + πΌ3 π 3 − π2 = 0 π 3 = 35.0 Ω πΌ3 These are all the equations we need to start solving Example - Kirchhoff's Rules Determine the magnitudes and directions of the currents in each resistor using Kirchhoff ’s rules π΄: πΌ2 = πΌ1 + πΌ3 1: π1 − πΌ1 π 1 − πΌ2 π 2 = 0 π = 25.0 Ω B: πΌ1 +πΌ3 = πΌ2 πΌ1 1 2: −π2 + πΌ3 π 3 + πΌ2 π 2 = 0 πΌ1 = πΌ2 − πΌ3 π1 = 9.0 π π 2 = 68.0 Ω π΄ 1: π1 − πΌ2 − πΌ3 π 1 − πΌ2 π 2 = 0 π΅ π1 = −π 1 πΌ3 + π 1 + π 2 πΌ2 πΌ2 π2 = 12.0 π π 3 = 35.0 Ω πΌ3 3: π1 − πΌ1 π 1 + πΌ3 π 3 − π2 = 0 9 π = −25πΌ3 + 93πΌ2 9 + 25πΌ3 πΌ2 = 93 Example - Kirchhoff's Rules Determine the magnitudes and directions of the currents in each resistor using Kirchhoff ’s rules π΄: πΌ2 = πΌ1 + πΌ3 1: π1 − πΌ1 π 1 − πΌ2 π 2 = 0 π = 25.0 Ω B: πΌ1 +πΌ3 = πΌ2 πΌ1 1 2: −π2 + πΌ3 π 3 + πΌ2 π 2 = 0 πΌ1 = πΌ2 − πΌ3 π1 = 9.0 π π 2 = 68.0 Ω π΄ 3: π1 − πΌ1 π 1 + πΌ3 π 3 − π2 = 0 2: −π2 + πΌ3 π 3 + πΌ2 π 2 = 0 π΅ π2 = πΌ2 π 2 + πΌ3 π 3 πΌ2 π2 = 12.0 π π 3 = 35.0 Ω 12 = 68πΌ2 + 35πΌ3 12 = 68 πΌ3 9 + 25πΌ3 + 35πΌ3 93 12 = 6.58 + 18.3πΌ3 + 35πΌ3 πΌ2 = 9 + 25πΌ3 93 Example - Kirchhoff's Rules Determine the magnitudes and directions of the currents in each resistor using Kirchhoff ’s rules π΄: πΌ2 = πΌ1 + πΌ3 1: π1 − πΌ1 π 1 − πΌ2 π 2 = 0 π = 25.0 Ω B: πΌ1 +πΌ3 = πΌ2 πΌ1 1 2: −π2 + πΌ3 π 3 + πΌ2 π 2 = 0 πΌ1 = πΌ2 − πΌ3 π1 = 9.0 π 12 = 6.58 + 18.3πΌ3 + 35πΌ3 π 2 = 68.0 Ω π΄ 12 = 6.58 + 53.3πΌ3 π΅ πΌ2 π2 = 12.0 π 3: π1 − πΌ1 π 1 + πΌ3 π 3 − π2 = 0 π 3 = 35.0 Ω 5.42 = πΌ3 53.3 πΌ3 = 0.10 π΄ πΌ3 9 + 25(0.10) πΌ2 = 93 πΌ2 = 0.12 π΄ πΌ2 = 9 + 25πΌ3 93 Example - Kirchhoff's Rules Determine the magnitudes and directions of the currents in each resistor using Kirchhoff ’s rules π΄: πΌ2 = πΌ1 + πΌ3 1: π1 − πΌ1 π 1 − πΌ2 π 2 = 0 π = 25.0 Ω B: πΌ1 +πΌ3 = πΌ2 πΌ1 1 2: −π2 + πΌ3 π 3 + πΌ2 π 2 = 0 3: π1 − πΌ1 π 1 + πΌ3 π 3 − π2 = 0 π1 = 9.0 π πΌ1 = πΌ2 − πΌ3 π 2 = 68.0 Ω π΄ πΌ1 = 0.12 π΄ − 0.10 π΄ π΅ πΌ2 π2 = 12.0 π πΌ1 = 0.02 π΄ π 3 = 35.0 Ω πΌ3 πΌ2 = 0.12 π΄ πΌ3 = 0.10 π΄ Batteries in Series and Parallel Our last example showed us that we can have multiple batteries in a circuit When two batteries are in series, the resulting voltage is the algebraic sum of the two batteries π1 = 9.0 π π2 = 12.0 π ππ‘ππ‘ = 21.0 π If the two batteries point in opposite directions π1 = 9.0 π π2 = 12.0 π ππ‘ππ‘ = 3.0 π If we think of this in terms of current, in this diagram current moves from low potential to high potential (gaining voltage) and as it cross the second battery it moves from high to low potential (losing voltage) Circuits – Capacitors in Series So far we have only discussed resistors, but capacitors are equally as common in electrical devices If we apply the loop rule, we can add up the voltage drops across each capacitor ππ‘ππ‘ = π1 + π2 + π3 πΆ3 π π π π = + + πΆπ‘ππ‘ πΆ1 πΆ2 πΆ3 1 1 1 1 = + + +β― πΆπ‘ππ‘ πΆ1 πΆ2 πΆ3 πΆ1 πΆ2 Circuits – Capacitors in Parallel When capacitors are in parallel, the junction rule will show πΌπ‘ππ‘ = πΌ1 + πΌ2 + πΌ3 ππ‘ππ‘ π1 π2 π3 = + + π‘ π‘ π‘ π‘ ππ‘ππ‘ = π1 + π2 + π3 πΆπ‘ππ‘ π = πΆ1 π + πΆ2 π + πΆ3 π πΆπ‘ππ‘ = πΆ1 + πΆ2 + πΆ3 + β― πΆ1 πΆ2 πΆ3 Electromotive Force A device like a battery that supplies a constant source of voltage is called a source of electromotive force (emf) The emf β° is a voltage supplied by the battery, and this is how we often label the voltage on a diagram. This is different than the terminal voltage which is the voltage across the whole circuit If you read that your AA battery has 1.5 π, this is really the emf that it is giving you The emf is not perfect though because there is some internal resistance across the battery itself. This internal resistance can change the Voltage you read across the battery depending on the amount of current passing through the source π = β° − πΌπ Measuring the voltage across the battery with no current reads the emf. When the battery is connected to a circuit, the current interacts with the internal resistance of the battery and lowers the voltage across it β° Example – Battery with Internal Resistance A 65.0 Ω resistor is connected to the terminals of a battery whose emf is 12.0 π and whose internal resistance is 0.5 Ω. β° = 12 π, π = 0.5 Ω a) Calculate the current in the circuit π = β° − πΌπ Circuit Side Battery Side We can apply Ohm’s Law to the voltage on the RHS π = 65.0 Ω πΌπ = β° − πΌπ From here, we can attempt to isolate the current πΌ πΌπ + πΌπ = β° πΌ= β° π +π πΌ(π + π) = β° πΌ= 12.0 π 65.0 Ω + 0.5 Ω πΌ = 0.183 π΄ Example – Battery with Internal Resistance A 65.0 Ω resistor is connected to the terminals of a battery whose emf is 12.0 π and whose internal resistance is 0.5 Ω. b) Calculate the terminal voltage of the battery β° = 12 π, π = 0.5 Ω π = β° − πΌπ π = 12.0 π − 0.183 π΄ 0.5 Ω π = 11.9 π π = 65.0 Ω Example – Battery with Internal Resistance A 65.0 Ω resistor is connected to the terminals of a battery whose emf is 12.0 π and whose internal resistance is 0.5 Ω. c) Calculate the power dissipated across the resistor π and the battery’s internal resistance π β° = 12 π, π = 0.5 Ω π = πΌπ 2 π = 0.183 π΄ 65.0 Ω 2 π = 2.18 π π = 65.0 Ω π = πΌπ 2 π = 0.183 π΄ 0.5 Ω π = 0.02 π 2 This is why batteries get hot when then supply current RC Circuits Finally, we want to look at combinations of resistors and capacitors in a circuit together When a voltage is applied across an RC circuit charge begins to accumulate across the capacitor and the potential across it begins to increase As voltage increases, the current slows down until the capacitor has the same voltage as the power supply where there is no more current flow + − As the capacitor is charging, the voltage across the capacitor takes an exponential shape in the form ππΆ = β° 1 − π −π‘/π Where π = π πΆ is called the time constant and dictates how fast a capacitor is charged If the power supply is cut off, then the capacitor begins to discharge and contribute to current in the circuit. The discharge potential looks like ππΆ = β°π −π‘/π − + Example - RC Circuits The capacitance in a circuit is πΆ = 0.30 ππΉ, the total resistance is 20 πΩ, and the battery emf is 12 π. Determine, a) the time constant π = π πΆ + − π = 20 × 103 Ω 0.30 × 10−6 πΉ π = 6.0 × 10−3 π − + Example - RC Circuits The capacitance in a circuit is πΆ = 0.30 ππΉ, the total resistance is 20 πΩ, and the battery emf is 12 π. Determine, b) the maximum charge the capacitor could acquire π = πΆβ° + − Remember β° = π − πΌπ, so the maximum value is when current is zero π = 0.30 × 10−6 πΉ 12 π π = 3.6 ππΆ − + Example - RC Circuits The capacitance in a circuit is πΆ = 0.30 ππΉ, the total resistance is 20 πΩ, and the battery emf is 12 π. Determine, b) the time it takes for the charge to reach 99% of this value This can be related to voltage across a capacitor which takes time to charge + − π = πΆβ° = πΆβ° 1 − π −π‘/π 0.99 πΆβ° = πΆβ° 1 − π −π‘/π 1 − 0.99 = 0.01 = π‘ − = ln 0.01 π π‘ = −π ln 0.01 π −π‘/π π‘ = − 6.0 × 10−3 π ln 0.01 π‘ = 28 × 10−3 π − + Multimeters When we want to measure the voltage or current, we have devices to do this. When combined into one device it is called a multimeter The multimeter has a knob that can be used to switch between different values that we want to measure The multimeter has two ports, a red side (high potential) and a com side (low potential) One wire is plugged into each port, and each is used to make a measurement When measure voltage, this is called a voltmeter When measure current, this is called an ammeter Voltmeter When measuring voltage, the voltmeter must be put in parallel π The resistance in a voltmeter is very high so that current does not pass through the voltmeter Voltmeter When measuring current, the ammeter must be put in series π΄ To do this we must break the circuit. This makes sure that the current flows into the ammeter Ammeter’s have very low resistance to ensure current runs through it Caution: Putting an ammeter in parallel will overload the ammeter which WILL break it
