Unit 3. Electrical Circuits

Strand G. Electricity Unit 3. Electrical Circuits Contents Page Representing Direct Current Circuits 2 Rules for Series Circuits 5 Rules for Parallel Circuits 9 Circuit Calculations 14

G.3.1. Representing Direct Current Circuits A direct current, abbreviated to D.C, is a constant charge flow in one direction. In a D.C circuit the flowing charges never reverse direction. This is the type of current obtained from all batteries, as well as D.C. power supplies. If your device needs charging, or runs on batteries, it is D.C, even Cell though it might be plugged into the mains supply to (a) charge (the mains supply is not D.C). To represent a D.C. circuit, the wire in which the electrons flow is represented by the circuit loop as shown in Figure G.3.1 (black lines). Note that the circuit MUST be a continuous loop. If there are any breaks in the circuit, charge will not flow. This is how the electrical switch works. When the switch is open it forms a gap or a break in the circuit loop and the charge won t flow (zero current). When the switch is closed it completes the circuit and charge can flow (non-zero current). Circuit Loop Within the circuit loop, a battery or power supply is required to provide the electrons with electrical potential and push them around the circuit (supplying the voltage and driving the current). In a D.C circuit, the I energy source may be called a battery or cell (a battery contains two or more cells) and is represented by two parallel lines, with one shorter than the other as shown Figure G.3.1.1 by Figure G.3.1.1. (a) and (b). The shorter line is always the negative (-) terminal whist the longer line is always the positive (+) terminal. The voltage of the cell is always stated, with the unit of volt V as shown. From electrostatics, likes repel and opposites attract. Therefore, electrons, which are negatively charged, flow from the negative terminal toward the positive terminal. Since conventional current takes the direction of positive charge flow, the conventional current in a circuit is always in the opposite direction to charge flow, and therefore from the positive to the negative terminal of the cell, as shown by the red loop in Figure G.3.1.1. (c). Components are devices in a circuit on which the flowing electrons do work. Electrical components include the cell, resistors, lamps, buzzers, diodes, voltmeters and ammeters. Each component has a particular use, and a specific circuit symbol. Some of the most common devices are listed below, including a description of the device and the corresponding circuit symbol. (b) (c) + + 6V - - 2

Circuit Symbol Component Description V Cell The source of electric potential in a circuit, providing the electric field required to accelerate the charges and produce a current. The longer parallel line is the positive terminal, and current flows from positive to negative. Lamp Contains a high resistance filament which when heated by the current passing through it, produces light. 5Ω Resistor A fixed resistor has an unchanging value of resistance usually stated on the circuit symbol. Reduces or limits the current in a circuit. V A Variable Resistor Voltmeter Ammeter LDR (Light Dependent Resistor) Switch Diode A resistor that allows the resistance to be manually varied. A Voltmeter is a device that measures potential difference between two points in a circuit. An ammeter is a device that measures the current flowing between two points in a circuit. An LDR is a variable resistor whose resistance varies as a function of the intensity of light falling upon it. A device that can form an intentional break in a circuit, allowing current to be switched on or off. A device that allows current flow in one direction only. LED (Light Emitting Diode) A device that allows current flow in one direction only, and emits light when a current passes through it (but does not rely on a high resistance filament). Buzzer Heater A device that vibrates when a current passes through it, producing sound. A device designed to transfer electrical energy to the surroundings via heat. 3

Worked Example. A student wants to measure the resistance of an unknown wire. He knows that if he knows that since V=IR, a plot of V vs. I would provide him with a gradient equal to the resistance that he could then calculate. Suggest a circuit that would allow the student to obtain the results he needs. Answer The student needs to be able to measure the voltage across the wire and therefore needs a voltmeter. He also needs an ammeter to measure the current in the circuit, allowing him to then plot his measurements on a graph of V vs. I. However, when combined with a cell or battery this setup will only allow one value of V and I to be obtained. In order to vary I and thereby obtain a spread of values for V, the student will also need a variable resistor. A suitable circuit would therefore be; V A V Unknown Wire 4

G.3.2. Rules for Series Circuits A series circuit is one in which all components are connected one after the other on the same loop, as shown by Figure G.3.2.1 (a). For components in a series circuit the electrons only have one path that they can take (Figure G.3.2.1 (b)). In addition, components do not use up charge. Therefore the charge per second flowing into a component is always the same as the charge per second flowing out of the component. Hence the flow of charge, or the current, must be the same at any point in a series circuit and the same through each component. (a) (b) 3V 3V 5Ω In a series circuit, the current flowing through each component is the same. In addition, the current measured anywhere in the circuit will be the same. The p.d, between any two points in the circuit is the energy transferred per Coulomb of charge as the charge flows from one point to another. As the charge flows through a component, they do work and the p.d across the component drops. As the charge goes through the cell or battery, the cell does work on the charges, and the p.d rises by an amount equal to the potential across the cells terminals. The p.d supplied by the cell is shared across all components within a series circuit. If we consider Figure G.3.2.2, the cell provides a voltage V0. The voltage across the individual resistors must therefore add up to V0. Electrons 5Ω Figure G.3.2.1 V0 V1 For two or more components in series, the total p.d across all the components is equal to the sum of the potential differences across each component. It therefore follows that; V0 = V1 + V2 + V3 V2 V3 V0 = V1 + V2 + V3 Figure G.3.2.2 For any complete loop, the sum of the emf s equals the sum of the p.d s round the loop 5

Worked Example A 6V battery is connected in series with a variable resistor, and a 2.5V 0.5W bulb. (a) Sketch the circuit (b) If the variable resistor is adjusted so that the p.d across the bulb is 2.5V, calculate the p.d across the resistor and the current through the bulb. (c) The resistance of the resistor. Answer (a) 6V (b) The variable resistor is adjusted such that the p.d across the bulb is 2.5V. The emf supplied by the battery is 6V. Since the sum of the emf s must be equal to the sum of the p.d s, 6V - 2.5V = 4.5V Therefore, the p.d across the variable resistor is 4.5V. Therefore, every Coulomb of charge leaves the battery with 6J of potential. 4.5J is delivered to the resistor and 2.5J to the bulb. The bulb is 0.5W, or 0.5J per second. Therefore 2 coulombs of charge pass through the bulb per second. Since I = Q t and t = 1, I = Q/1 = 2A. (c) Since I through the bulb is 2A and the current is the same everywhere in a series circuit, the current through the resistor is also 2A. Using Ohms Law, RR = VV II = 4.5VV 2AA = 2.25Ω 6

For resistors in a series circuit, just as any other component in a series circuit, the current is the same through each resistor and the total p.d of the combined resistors is equal to the sum of the individual p.d s. If we consider these two rules together VT with Ohm s Law for two resistors connected in series as shown by Figure G.3.2.3 we see that; I R1 R2 I For the p.d across R1, V = I R1 For the p.d across R2, V = I R2 V1 Figure G.3.2.3 V2 The total p.d (VT) across both resistors is therefore VT = I R1 + I R2 and the total resistance across the two resistors is RR TT = VV TT II = IIII 1 + IIII 2 = RR II 1 + RR 2 giving the general result for resistors in series; For two or more resistors connected in series, the total resistance is equal to the sum of the individual resistances. RT = R1 + R2 +R3 + Worked Example A 9V battery is connected to a series circuit containing a 6Ω and 4Ω resistor. Calculate the total resistance, and the current through the 6Ω resistor. Answer Since the circuit is a series circuit, the individual resistances of the resistors may be combined and considered as a total resistance using RT = R1 + R2 RR TT = 6 + 4 = 10Ω The current is the same everywhere; so using Ohms Law to calculate the current around the whole circuit provides us with the current through the 6Ω resistor; II = VV RR = 9VV 10Ω = 0.9AA 7

Exercise G.3.2 1. A, 4Ω, 7Ω, and 6Ω resistor are connected in series on a single loop of wire with a 9V battery. Calculate the total resistance in the circuit and the current drawn from the battery. What would be the current drawn if the circuit resistance was halved? 2. Two resistors are connected in series with a cell. The p.d across the first resistor is 6V and the p.d across the second resistor is 4.5V. Calculate the emf supplied by the cell. 3. A 2Ω and a resistor are connected in series with a 6V battery. Calculate the current through the 2Ω and then the resistor. 4. 3 bulbs are connected in series with a 12V battery. The p.d across the first bulb is 3V. The p.d across the second is 6V. Draw the circuit, and calculate the p.d across the 3 rd bulb. Challenge Question 5. A 12V battery is connected in series with an ammeter, a 20Ω resistor, and an unknown resistor R. (a) Sketch the circuit diagram (b) The ammeter reads 0.4A. Calculate (i) The p.d across the 20Ω resistor (ii) The p.d across R (iii) The resistance of R 8

G.3.3. Rules for Parallel Circuits 6V Parallel circuits, such as the circuit shown in Figure G.3.3.1 consist of two or more components connected to a cell in separate branches, similar to a ladder 1 arrangement, providing two or more pathways that 6Ω the current can follow. This type of arrangement has some advantages over the series circuit. For example, 2 the electric lights of your home are wired on a parallel circuit. If they were wired in series, if one 4Ω light bulb blew, this would form a break in the entire circuit and no current would flow. All the lights in the Figure G.3.3.1 house would go out! Instead, on a parallel circuit, the current can take an alternative route around the circuit, and only the lights on that branch of the ladder would go out until the bulb has been changed. When an electron leaves the cell and reaches the first junction (red dot, Figure G.3.3.1) it is presented with a choice. It may pass along the branch containing the 6Ω resistor and return to the battery, following current loop 1 (marked in the direction of electron flow), or it may continue around loop 2, passing through the 4Ω resistor and returning to the battery. This choice at a circuit junction or node leads us to Kirchoff s First Law; At any junction in a circuit, the sum of the currents entering the junction is equal to the sum of the currents leaving the junction. This is actually a re-statement of the conservation of charge. Worked example. Consider the following circuit junctions and calculate the unknown current I in each case. Hint: Pay careful attention to the direction of the current I = 3A Answer I = 0.5A (a) I =? I =? I = 1.5A I = 2A (b) For junction (a) we have 0.5A entering the junction and 3A leaving the junction. Since the sum of the current entering must equal the sum of the current leaving, the unknown current is; 3A 0.5A = 2.5A entering the junction. 9

For junction (b) we have 1.5A and 2A with both known currents entering the junction. Therefore, the unknown current is; 1.5A + 2A = 3.5A leaving the junction. Considering Figure G.3.3.1 once more, the electrons arriving at the junction either pass through the 6Ω resistor, or through the 4Ω resistor. Since each electron carries the same amount of electric potential, delivering the same amount of energy from the cell whichever resistor it goes through, the potential difference across each resistor is the same. For components connected in parallel, the potential difference across each component is the same. Since components in parallel have identical potential difference across them, the current passing through each component depends on the resistance. Since from Ohm s Law II = VV RR the greater the resistance the smaller the current. Calculating the current through a component in a branch thus allows the current exiting the junction along that branch to be obtained. Worked Example. Consider the following circuit and calculate the current through each individual resistor, and the current through the battery. Answer Here the battery supplies 6V. Since in a parallel circuit the p.d is the same across all components, each resistor in the circuit has a p.d of 6V. We can therefore calculate the current through each resistor using the individual resistances; II 2 = VV RR 1 = 6VV 6Ω = 1AA II 4 = VV RR 2 = 6VV 4Ω = 1.5AA II 5 = VV RR 3 = 6VV 2Ω = 3AA I1 I3 I2 I4 I5 6V 6Ω 4Ω 2Ω I6 10

Since the sum of the currents leaving a junction equals the sum of the currents entering, I1 = I2 + I4 + I5 = 1A + 1.5A + 3A = 5.5A Similarly, and I3 = I1 I2 = 5.5A 1A = 4.5A I6 = I5 + I4 = 3A + 1.5A = 4.5A Because resistors in parallel all have the same p.d, and the current through a parallel combination of resistors is equal to the sum of the individual currents, for the two resistors R1 and R2 as shown in Figure G.3.3.2, I1 R1 II 1 = VV RR 1 aaaaaa II 2 = VV RR 2 I V I The total current through the resistor combination in parallel is therefore I2 R2 Figure G.3.3.2 II TT = II 1 + II 2 = VV RR 1 + VV RR 2 Since the total resistance RT = V/IT, VV and cancelling V we obtain RR TT = VV RR 1 + VV RR 2 1 = 1 + 1 RR TT RR 1 RR 2 and the general result For two or more resistors in parallel, the total resistance RT is given by; 1 RR TT = 1 RR 1 + 1 RR 2 + 1 RR 3 + If you are used to adding fractions, you would have already noted that, for resistors in parallel, the total resistance is ALWAYS less than the smallest of the individual resistances. 6V 2Ω 4Ω 6Ω 11

Worked Example. Consider the following diagram and calculate the total current drawn from the battery. Answer To use Ohm s Law to calculate the current drawn from the battery, the total circuit resistance is required. Here it is easiest to calculate the total resistance of the three parallel resistors first, and then replace them with a single resistance value that is in series with the resistor. The total resistance of the parallel circuit (inside the red dashed box) is given by; 1 = 1 RR TT 2 + 1 4 + 1 6 = 11 12 = 0.92 Therefore, RT = 1.09Ω (don t forget that it is 1/RT calculated above, and so we have to take the inverse to find RT). Now the parallel circuit can be replaced as shown with a single resistor of resistance RT. Thus we obtain a and a 1.09Ω resistor connected in series. Since for resistors in series, RT = R1 + R2, the circuit resistance Rc is given by Rc = + 1.09Ω = 4.09Ω Then the current drawn from the battery is II = VV RR = 6VV 4.09Ω = 1.47AA 2Ω 4Ω 6Ω 6V 1.09Ω Exercise G.3.3 1. Consider the following current junctions and calculate the current and current direction in each case. I = 0.5A (a) I =? I = 4A (b) I = 2A I = 1.5A I = 3A I =? I =? I = 3.5A I = 1.5A (c) 12

2. Three fixed resistors, a 2Ω, 4Ω and 8Ω, are connected in parallel with two cells of 3V each. The cells are connected in series, beside each other. Calculate the potential difference across each resistor. 3. Draw a circuit diagram of the circuit described in question 2 above. Calculate the current across each resistor. 4. Without calculation, state the maximum value of the combined resistance for each of the following resistor combinations. Then calculate the exact value of the combined resistance for each. 2Ω 2Ω 4Ω (a) (b) 5Ω (c) 4Ω 8Ω 5Ω 2Ω (d) Challenge Question 5. Consider the following circuit and calculate the total current drawn from the battery. 9V 4Ω 6Ω 12Ω 13

G.3.4. Circuit Calculations For circuit calculations (which can get quite complicated) it is useful to put together a problem solving strategy. Review relevant concepts This encompasses the majority of concepts introduced in this strand, summarised as follows; II = QQ tt Current flows from the positive to negative terminal, electrons flow from the negative to positive terminal Potential difference (p.d) or voltage (V) is the work done (or energy transferred) per unit charge. Work W is done when a charge Q flows through a component VV = WW QQ The emf (ε) is the energy per unit charge supplied to the circuit. The potential difference (p.d) or voltage is the energy used by, or the work done on components in the circuit. Electrical Power P is given by; PP = IIII Ohm s Law states; V = IR In a series circuit, the current flowing through each component is the same. At any junction in a circuit, the sum of the currents entering the junction is equal to the sum of the currents leaving the junction. For any complete loop, the sum of the emf s equals the sum of the p.d s round the loop For two or more components in series, V0 = V1 + V2 + V3 For components connected in parallel, the potential difference across each component is the same. For resistors in series, RT = R1 + R2 +R3 + 14

For resistors in parallel, 1 RR TT = 1 RR 1 + 1 RR 2 + 1 RR 3 + Set Up the Problem 1. Make a drawing of the circuit. This will help to split the circuit up into easier chunks in order to solve. 2. Identify all components. 3. Identify sources of emf and of p.d so that concepts such as the sum of the emf equals the sum of the p.d on a closed loop, or the p.d of resistors in parallel equal the emf of the source may be used. 4. Identify the required quantity Solve the Problem 1. Simplify where possible. 2. Combine parallel resistors into a combined resistance 3. Combine series resistors in complex problems 4. Calculate total resistance of the circuit to find current drawn 5. Employ the junction rule to find the current in each branch Remember that a diode only passes current in one direction. Also, for more than one cell in series, the individual emf s may be added. Evaluate There are always different ways to solve a problem. For instance, junction rules could be used to solve for currents in a parallel branch, but also the individual resistance could be used along with the source voltage. Check your answers by solving both ways. Also remember the combined resistance of a set of parallel resistors can never be greater than the smallest resistance value in the combination. And don t forget to take the inverse when calculating the resistance of parallel resistors! 15

Worked example. Consider the following circuit and calculate the current through the 6Ω resistor, and the 20Ω resistor. Then calculate the power through the 6Ω resistor. 20Ω 15Ω 30Ω 30Ω 60Ω 6Ω Answer. 6V 3V First we review the concepts. We want power, which is given by P = I V, and for this we need the current through the resistor and the p.d across the resistor, and we need to calculate the current in individual branches. In addition, the two cells are separate, but we can sum these to a single emf of 9V. Let s now simplify the problem by splitting the circuit up into separate current loops. Loop 3 (the whole circuit) Loop 1 Loop 2 20Ω 15Ω 30Ω 30Ω 60Ω 6Ω 9V First we look for the total current in the circuit since this is the current through the 6Ω resistor. To find this we want the total resistance of the circuit. Starting with loop 1: 1 RR LL1 = 1 20ΩΩ + 1 30ΩΩ = 1 12ΩΩ ttheeeeeeeeeeeeee RR LL1 = 12ΩΩ 16

Now loop 2. First we combine the two series resistors on the top branch R = 15Ω + 30Ω = 45Ω Then the combined resistance of loop 2 is 1 = 1 RR LL2 45ΩΩ + 1 60ΩΩ = 7 180 ttheeeeeeeeeeeeee RR LL1 = 180 7 = 25.7ΩΩ We now have a much simpler loop 3. Let s redraw it Loop 3 (the whole circuit) 12Ω 25.7Ω 9V 6Ω Now we can easily calculate the total resistance of the circuit using the series combination rules. RR TT = 12ΩΩ + 25.7ΩΩ + 6ΩΩ = 43.7ΩΩ And the total current from the cell combination using Ohm s law. II TT = VV RR TT = 9VV 43.7ΩΩ = 0.21AA All of this current flows through the 6Ω resistor. The power through the resistor is found using P = IV, so we now need the p.d across the 6Ω resistor. Now we know the current through the resistor we can use Ohms law once more; VV 6ΩΩ = IIII = 0.21AA 6ΩΩ = 1.26VV And the power through the resistor is therefore 17

PP = IIII = 0.21AA 1.26VV = 0.26WW Let s now turn our attention to the 20Ω resistor of loop 1. We know that the current flows from positive (the long line of the cell) to the negative (short line of the cell). The current therefore enters as shown. There are no junctions between loop 1 and the battery so this is the total current I = 0.21A. The current is shared across the two resistors, the size of the current through each being dictated by the I Loop 1 20Ω 30Ω size of the resistance. Since the 20Ω resistor is 2/5 th the size of the 30Ω resistor, 2/5 th of the current goes to the 20Ω resistor and 3/5 th to the 30Ω resistor. Thus Exercise 3.4 2 0.21AA = 0.084AA 5 1. Consider the following circuit. If each bulb has a resistance of 25Ω, what will the reading be on the ammeter? 10Ω 25Ω 30Ω A 35Ω 50Ω 12V 2. In the following circuit, each bulb has a resistance of 20Ω and the variable resistor is set to 30Ω. Calculate the reading on the voltmeter (a) with the switch open, and (b) with the switch closed. 12V 25Ω V 50Ω 30Ω 18

3. For the following circuit the bulb resistance is 5Ω and the variable resistor is set to 6Ω. Calculate the unknown currents I1, I2 and I3. 12V I1 A I2 6Ω 4Ω D B C I3 4. Complete the table for the following circuit. R1 R2 Total Voltage Current Resistance 2k 6k Power 24V R1 2kΩ R2 6kΩ Challenge Question 5. An A level physics student sets up the following circuit, and sets the variable resistor to 0.5Ω. Calculate the current entering and leaving junction 1 (j1) and the heat energy dissipated by the heating element after 4 minutes. Assume the diode has negligible resistance. The student then reverses the battery terminals causing the current to flow in the opposite direction. Recalculate your values. j1 15Ω 12V 0.4Ω 2Ω 19