## Terminal Voltage All voltage sources have two parts – a source of electrical energy that has an [[Circuit Theory (1)#Electromotive Force|electromotive force]] (EMF), and an internal [[Circuit Theory (1)#Resistance|resistance]] $r$. ![[Circuit Theory (1)#Electromotive Force]] EMF is the potential difference of a source, measured across the terminals when **no current is flowing**. The internal resistance $r$ affects the output voltage when there is a current ($V=IR$). **Terminal voltage** ($V$) is the potential difference across the positive and negative terminals of a battery. When multiple voltage sources are in series, their internal resistances add and their EMFs sum algebraically. $ V = EMF-Ir $ > [!TIP] When there is no current $I=0$ and $EMF = V$. When there is current, $V < EMF$ ## Maximum Power Transfer A *load* in electrical terms is any component added to a circuit which consumes power (e.g a resistor). We can represent any load as a resistor having resistance $R_{l}\ \Omega$. The **maximum power transfer theorem** states that: > DC voltage source will deliver maximum power only when the load resistance $R_{l}$ is equal to the source resistance $R_{s}$. * When $R_{l}$ is too small: more current flows, but most voltage drops across the source resistance so the load gets less power * When $R_{l}$ is too large: more voltage appears across the load, but very little current flows so again the load gets less power > [!WARNING] Practical application > At maximum power transfer ($R_{l}=R_{s}$) **the efficiency is only 50%** because half the power is dissipated in the source resistance ### But why? Consider voltage source $V$, internal resistance $R_{s}$ and load resistance $R_{l}$. The current flowing through the circuit is $I=\frac{V}{R_{s}+R_{l}}$. The power delivered to the load is $P=I^2 \times R_{l}$. Substitute the current: $ P=\frac{V^2 \times R_{l}}{(R_{s}+R_{l})^2} $ Using calculus, we can prove $P_{max}$ is reached when $R_{s}=R_{l}$. Take the derivative of $P$ with respect to $R_l$ and set it equal to zero: $ \frac{dP}{dR_l} = 0 $ Using the quotient rule: $ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} $ Where $u = V^2 R_L$ and $v = (R_s + R_L)^2$ $ \frac{dP}{dR_l} = \frac{(R_s + R_l)^2 \cdot V^2 - V^2 R_l \cdot 2(R_s + R_l)}{(R_s + R_l)^4} $ Simplifying the numerator: $ \frac{dP}{dR_l} = V^2 \cdot \frac{(R_s + R_l)^2 - 2R_l(R_s + R_l)}{(R_s + R_l)^4} $ Factor out $(R_s + R_l)$: $ \frac{dP}{dR_l} = V^2 \cdot \frac{(R_s + R_l) - 2R_l}{(R_s + R_l)^3} $ $ \frac{dP}{dR_l} = V^2 \cdot \frac{R_s - R_l}{(R_s + R_l)^3} $ Setting the derivative equal to zero: $ \frac{V^2(R_s - R_l)}{(R_s + R_l)^3} = 0 $ Since $V^2 \neq 0$ and $(R_s + R_L)^3 \neq 0$: $ R_s - R_l = 0 \qquad \boxed{R_{l}=R_{s}} $ ## Measuring Devices A multimeter is an instrument that can measure multiple electrical properties. There are three components: * Voltmeter for measuring voltage * Ohmmeter for resistance measurement * Ammeter for measuring current ### Ammeter The ammeter (also known as ampere meter) is used for measurement of current. It must be connected **in series**, breaking the circuit and inserting the ammeter in the gap. Connecting an ammeter adds resistance (from the device itself), which will change the current and affect your measurement. > [!NOTE] An ideal ammeter would have **zero resistance**, so it doesn't affect the current being measured Ammeters are protected from excessive current by a small fuse. If the ammeter is connected across a substantial voltage source, the resultant surge would blow the fuse. ### Voltmeter When using a voltmeter, the connection must be made **in parallel** with the part of the circuit across which the voltage measurement is to be taken (e.g across a resistor). Connecting a voltmeter also adds a resistance, which affects the circuit behaviour. > [!NOTE] An ideal voltmeter would have **infinite resistance**, so no current flows through it and it only measures voltage without changing the circuit Remember that resistors in parallel result in lower resistance than either node individually (see [[Circuit Theory (2)#Resistors in Parallel|resistors in parallel]]). This means when using a voltmeter the equivalent resistance is lower, resulting in more current flowing. Say you're measuring voltage across a $1M\Omega$ resistor and your voltmeter has $10M\Omega$ internal resistance. With the voltmeter in parallel, $R_{eff}=\frac{1\times10}{1+10}=0.91M\Omega$. ### Ohmmeter Resistance can only be accurately measured across components when they are **disconnected from the circuit**. If you attempt to measure the resistance across a component while in a circuit, you'll see the total effective resistance across all branches of the circuit connected to that component. > [!TIP] Calculate $R$ across a component by measuring $V$ and $I$, then use [[Circuit Theory (2)#Ohm's Law|Ohm's Law]] ## Wheatstone Bridge > [!ERROR] This was removed from the curriculum and won't be in the exam! A Wheatstone Bridge is an alternative method of measuring resistance by using two [[Circuit Theory (2)#Voltage Divider|voltage dividers]]: * One has two precisely known fixed resistors, $R_{1}$ and $R_{2}$ * The other has unknown resistor $R_{x}$ in series with an variable resistor $R_{s}$ ![[Wheatstone bridge.png]] By adjusting $R_{s}$ until $V_1$ and $V_{x}$ are **balanced**, we can determine $R_{x}$. To determine if the bridge is **balanced**, we measure the voltage $V_{out}$ between the two bridges (e.g with one probe between $R_{1}$ and $R_{2}$, and the other between $R_{x}$ and $R_{s}$). **The bridge is balanced if $V_{out}=0V$**. For a balanced bridge, the voltage across $R_{1}$ and $R_{2}$ is the same. This also applies to voltage across $R_{s}$ and $R_{x}$. Therefore: $ V_{R1}=V_{R2} \qquad V_{R3}=V_{R4} $ The ratios can be written as: $ \frac{V_{R1}}{V_{R3}}=\frac{V_{R2}}{V_{R4}} $ From Ohm's Law, we get: $ \frac{I_{1}R_{1}}{I_{3}R_{3}}=\frac{I_{2}R_{2}}{I_{4}R_{4}} $ Since $I_{1}=I_{3}$ and $I_{2}=I_{4}$ (from [[Circuit Theory (2)#Kirchhoff's Current Law|KCL]]) we get: $ \frac{R_{1}}{R_{3}}=\frac{R_{2}}{R_{4}} $ Any one of the resistors could be the unknown component. Let's call it $R_{x}$ and assume that it is $R_{1}$ in this circuit: $ \frac{R_{x}}{R_{3}}=\frac{R_{2}}{R_{4}} $ Since $\frac{R_{2}}{R_4}$ is fixed, we need to adjust $R_{3}$ as the variable resistor, $R_{s}$: $ R_{x}=R_{s}\times\frac{R_{2}}{R_4} $ The Wheatstone Bridge is often used with **transducers** to measure physical quantities like temperature, pressure and strain. They're commonly used in applications where small changes in resistance are to be measured in sensors. This is used to convert a change in resistance to a change in voltage of a transducer. For example, to measure/use the change in resistance of a thermistor (a type of resistor whose resistance is dependant on temperature) by converting it to a change in voltage. ## Kirchhoff's Current Law #circuit-theory/kcl We briefly touched on Kirchhoff's Current Law (KCL) in [[Circuit Theory (2)#Kirchhoff's Current Law|week 2]]. To recap: > The total current entering a circuits junction is exactly equal to the total current leaving the same junction In other words, $I_{t}=I_{1}+I_{2}+I_{3}+\dots$ ![[KCL T junction.png]] In the diagram below, let all the currents entering the junction be positive values, and all the currents leaving the junction be negative values. ![[KCL Y junction.png]] $ I_{1}+(-I_{2})+(-I_{3})=0 $ ## Kirchhoff's Voltage Law #circuit-theory/kvl KVL states that the sum of voltages in a loop is equal to the [[Circuit Theory (1)#Electromotive Force|EMF]] of that loop. Put formally: > In a closed loop, the algebraic sum of the product of current and resistance in each part of the circuit is equal to the resultant EMF in the loop The *product of current and resistance* is voltage. Translated, KVL simply states that the **sum of voltage drops around a closed loop is equal to the EMF**. We can express this as: $ EMF = IR_{1}+IR_{2} $ ![[KVL example.png]] Another way to look at it is the **sum of potential differences in any loop must equal 0V**. We can use this to get to [[Circuit Theory (2)#Ohm's Law|Ohm's Law]] from KVL. $ V_{R1}=I\times R_{1} \qquad V_{R2}=I\times R_{2} $ $ V_{s}+(-IR_{1})+(-IR_{2})=0 $ $ V_{s}=IR_{1}+IR_{2} $ $ V_{s}=I(R_{1}+R_{2}) $ $ V_{s}=IR_{t} $ ![[Circuit Theory (4)#Branch Current Method]]