Concepts of Electrical Principles

Concepts of galvanising PrinciplesEverything is made of atoms in turn atoms consist of a combination of minuscule particles k n accept as neutrons, protons and electrons. The core of an atom consists of protons and neutrons while electrons exist in a cloud surrounding and rotating nigh the nucleus. The electron and proton ar capable of holding an electrical charge electrons hold veto charges and protons positive charge. We retire that like charges labour to each(prenominal) one other while opposite charges study the opposite effect in attracting bingle another.If we call to measure the take to the woods of electrons around a go we refer to this as a measure of electrical flowing. Electric stream is represented by the figure I and is a metre of charge carriers short a utilisen point in a circle. This is calculated as vitamin C of charge deviance a defined point in nonp beil second, which as a social unit is given the address ampere abbreviated to A. This st inker be measured usage an instrument called an ammeter which when affiliated in serial publication with a enlistment to measure the accepted passing by means of it.For electric menses to flow around a locomote there moldiness be a voltage crossways it. electromotive force is a measure of the potentiality difference (p.d), which acts like electric force pushing the current around the circuit. The pressure put forward be read in a circuit by a voltmeter, which must(prenominal) be applied by the foe. This happens when there is a deficit of electrons in a conductive material and this is so machine-accessible to another material with excess electrons. This is the case in a battery where chemicals allow electrons to flow from the negative terminal that contains an excess of electrons and the positive terminal containing positively charged protons. This happens beca social function opposite charges attract cardinal another.1.4 ResistanceThis flow of current faces oppos ition from confrontation this is a quantity of how much the electrons bump against the specific conductor they are current through. Some materials conduct electricity better then others. Materials that shake up a high shelter conduct electricity less well. Resistance limits the flow of electrons between the positive and negative ends of a circuit. We measure resistance in units called ohms (). One ohm is defined as the amount of resistance you come in a conductor when applying one volt of electrical pressure creates one amp of current.1.5 brawninessWhen electrons sit high in there shells surrounding the nucleus they have electrical energy. This energy asshole be harnessed to do work in various ways, if the electrons bump into atoms this corporation ca affair them to hightail it around which creates heat, they create electromagnetic waves as they travel which prat use there attraction and inconsistency to move things magnetically, and if the electrons move down there elec tric shells they give up excess energy boastful out light in the manner of photons.1.6 Charge CarriersThe sub-atomic particles that carry charge are known as protons and electrons as forwardly discussed electrons are negatively charges while protons are positively charged. The unit to measure the quantity of electrical charge (Q) is the coulomb (C) where 1 coulomb of charge is equal to charged electrons. If one coulomb of charge passes a point in one second we say this is one ampere of current. We can use our knowledge of math to deduct that if then if we take (I) as the current in amperes and t as the time in seconds thenElectrical Principles/Kirchhoffs Laws2.1 Potential DifferenceThe pull created by the difference in charge between the both sides of a circuit is called the potential difference, which is otherwise known as the voltage. potential sources that have higher attractive forces are known to have a higher potential difference.The units we use to measure voltage/potentia l difference is known as the ampere which is explained in naval division 1.6 as one coulomb of charge passing a given point in one second.2.2 Ohms Lawa) Ohms fairness relates potential difference, Current and Resistance in the spare-time activity equationI = current in amperesV = voltage in voltsR = Resistance in OhmsThis law states that the current I flowing in a circuit is directly proportional to the voltage applied to it and inversely proportional to the resistance.b) For a 5m distance of wire with a resistance of 600 ohms we can apply this law. If you where to half the length of wire you would half the resistance as there would be half as much material for the electrons to bump into.c) If we where to attach the length of the wire to 8m we can see that the resistance increases as create more material for the electrons to smash into.d) To suffer the length of the similar wire when the resistance is 420 ohms we do the following midpointSo we can say that the same wire w ith a resistance of 420ohms would measure 3.5 meters.2.3 Resistance VariationIf a ingredient of wire has a cross sectional area of 2mm2 and a resistance of 300 ohmsFind the resistance of the same length of wire if the cross sectional area is 5mm2.Given that resistance is inversely proportional to cross sectional area, increasing the cross sectional area increases the flow of electrons, we can calculate this mathematically as suchb) Find the cross sectional area of a wire of the same length and material of resistance 750.2.04Calculate the resistance of a 2km length of aluminium overhead ply crease if the cross sectional area of the cable is 100mm2. Take the resistivity of aluminium to be 0.03 x 10-6 mShow the equation you are using in your answer.We know that and that if we combine these rules we can create the ordinance . With one more piece of information we exit be able to take the material employ into account.This is do by including the resistivity of the material into the relationship considering it as a constant of proportionality. We use the symbol (Greek rho). The final equation exit look like this2.5 PowerIf electrical energy (W) = Charge (Q) x Voltage (V) then -a) Show the equation for power in scathe of current( I) and voltage (V).Electrical susceptibility (W) = Charge (Q) x Voltage (V)W = Q x VPower (P) = Current (I) x Voltage (V)P= V x Ib) employ Ohms law explain how power can overly be expressed in terms of I and R, and, V and R.P= V2/RP = I2RC) An e.m.f. of 250V is connected across a circuit resistance and the electric current through the circuit resistance is 4A. What is the power troubled in the circuit?2.6a) To discover the potential difference across the winding we use Ohms law as followsVoltage (V) = Current (I) X Resistance (R)V= 5A X 100V = 500Vb) If we tender to find the power dissipated by that same gyre we use our equations for powerPower (W) = Voltage (V) X Current (I)P= 500V x 5AP = 2500 Watts2.7A 12V battery is conne cted a shoot having a resistance of 40.a) tempt the current flowing in the load.For this we must again use Ohms law rearranged to make I the subject.I = V /RI = 12V / 40I = 0.3 Amps watch over the power consumed by the load.To calculate this we use our power equation again using the figure we scantily calculated for the current.P = VIP = 12V x 0.3AP = 3.6 wattsc) Determine the electrical energy dissipated in 2 minutes.Electrical Energy (W) = Charge (Q) x Volts (V)Current is charge per second and we sight that this circuit runs 0.3Amps, finding how much energy is dissipated in 2mins first requires changing minutes to seconds.2mins = great hundred secondsW = Q x VW = (120 X 0.3) x 12VW = 432 Watts2.8a) Explain what is meant by one unit of electricity with reference to Electrical Charge (Q), Voltage (V) and Time (T).A standard unit of electricity is usually calculated as a Kilowatt-hour (KWh), Which is 1000 watts of electricity dissipated for one hour. recognise MY BOOK ON THISb) Determine the power dissipated by the element of an electric fire of resistance 20 when a current of 10A flows through it.For this situation we are provided with the current at 10A and the resistance at 20 therefore we can use our power equation to find how much power is dissipated.P = I2RP = 102 x 20P = 2000 wattsc) If the fire is on for 6 hours chequer the energy use and the cost if 1 unit of electricity costs 13p.Firstly we take the power consumption in watts from we determined in question b then apply the following equation to itCost per social unit x Watts / 1000Multiply the per-hour cost by the running time.26p x 6h = 1.56p2.9 go bad this resistors in serial circuita) Express V in terms of V1, V2 and V3.VT = V1 + V2 + V3Voltages in this circuit forget each have a different observe if the resistances are different but if you add all the values unitedly they should in good equal the supply voltage.b) Express the total circuit resistance (RT) in terms of R1, R2 and R3.Re sistances in series always add in concert.This can be expressed asRT = R1 + R2 + R3c) Express in terms of I what the electric current is through the ammeter-A, R1, R2 and R3.In a series circuit the current is the same in any part of the circuit so construes using the ammeter would be the same as any reading taken on each of the resisters R1, R2 or R3.2.10A 12V battery is connected across a circuit having three series-connected resistors of resistances 4, 9 and 11.a) Determine the electric current through the circuit.As this is a series circuit the current would be the same throughout the circuit, to calculate this we must use ohms law, first we know that resistances add together in a series circuit to give the resistance total.4 + 9 + 11 = RT = 24Then we must implement Ohms lawI =V/RI = 12V / 24I = 0.5Ab) Determine the p.d. across the 9 resistor.Via Ohms law and our previous current calculation, we calculate the voltage across the 9 resister.V2 = I x R1V2 = 0.5 x 9V2 = 4.5 Voltsc) Determine the power dissipated in the 11 resistor.P = I2RP3 = 0.52 x 11P3 = 2.75 W2.11Two resistors are connected in series across a 24V supply with a flow of electric current of 3A within the circuit. If one of the resistors has a resistance of 2 determinea) The value of the other resistor.R2 = RT R1R2 = 8 2R2 = 6Trusting in Ohms law we can find the value of the other resistor using the values given for total voltage and current and knowing that resistances in series add together to give the resistance total.RT = V/IRT = 24/3RT = 8b) The p.d. across the 2 resistor. lick this requires Ohms law.V1 = I x R1V1 = 3A x 2V1 = 6 Voltc) How much energy is used if the circuit is connected for 50 hours.P=VIP=24v x 3P=72W50h = 180000sW = Q x VQ (charge) = I (current) x t (time)W =180000 x 72W = 12960000 Watt/joules2.12Analyse the resistors in parallel circuit.a) In terms of V, express the p.d. across R1, R2, and R3.V= I1R1 = I2R2 = I3R3We see that the voltage is the same across each resistor .b) Express the total load current I in terms of I1, I2, and I3.2.13For the circuit shown below, determinea) The reading on the ammeter,In a purely parallel circuit the voltage get out be the same in each branch of the circuit.V=I x RV = I1 x R1V = 8 x 5 = 40VI = V/R3I = 40/20 = 2Ab) The value of resistor R2.We now have all the values for I,= 11 8 2 = 1AR2 = V/I2R2 = 40/1R2 = 402.14Find the value of resistor that can replace the six resistors in this diagram.We know that resistances in series can be added together to give the total resistance, in this utilisation we have a parallel internet of resisters in series with 3 more resisters. Treating this parallel network as a single resistance go out allow us to calculate the total resistance of the circuit easily.Convert the resistances to conductanceAdding them together gives us the total conductance0.52GThis can then easily be reborn to resistance.Now the parallel circuit can be treated as a single resister, we can add all the resistors together and find the total resistance of the circuit giving us the value of a resister we can replace it with.2.15Analyse the circuit below and determineThe currents I1, I2, I3, I4, I5, and I6We can treat the two fastens of parallel resisters as single resisters if we first convert them to conductance and then for each add the conductances together then convert back to resistance.For the set of 3 parallel resistersThe Set of twoThe three resisters can be added to give our RTWe can now add these conductances together giving us our total conductance for the set of two resistors. This can then be converted to a combined resistance easilyWe now proceed to do this for the set of three resistorsWe now have the equivalent of 3 resistors in series, which we know can be added together to create a single resistanceNow that we know the total resistance for the circuit we can find I1 easily using Ohms lawWe must now find the voltages V1, V2 and V3 in order to later find the currents through the network branches.=20VAnd now V2Next I will calculate V3We can check this by adding all of the voltages to see if they equal the total voltage we have been given.This is over by 1.4V but I believe this is due to the compound effects of the rounding card and that the calculations made are correct.We know that the current through I1 is 5A now we will work out the currents through the branches of the parallel resistances using Ohms law2.16State Kirchoffs first (current) law. Show that the currents I2 and I3 combined are equal to the foreplay current I1Kirchhoffs Current Law statesThe sum of the currents entering a particular point must be zero.So all currents entering a point must equal all the currents flowing from it. thusly we must now think of the currents flowing from the junction as negative currents.i1+i2+i3+i4= 0Observing our circuit we see 11A of current going in, this means that the same amount of current must come out. thenceTo prove this we calculate I1 and I2 using Ohms lawI2= V/RI2=10/10I2= 1AI3= V/RI3=10/1I3= 10AWe can now calculate I1 expecting it to equal our given figure of 11A.I1= I2 + I3I1=10+1I1=11A2.17Using Kirchhoffs first (current) law, calculate current I1 and I2 in the network below.Kirchhoffs first current law states that the sum of the current entering a point must be zero.Examining the junctions we have 1.2A and 4.5A flowing in and 0.6A and I1 are flowing out.1.2 +4.5 = I1+0.61.2 + 4.5 0.6 = I1I1= 5.1AFor I2 there are three currents flowing in but none flowing out. This must mean that the last value is a negative value.5.1+3 + I3 = 08.1 + I3 =0I3 = 8.1A2.18The potential splitter shown below is used as a simple voltage calibrator. Determine the outturn voltage produced by the circuit(a) When the output terminals are left open-circuit (i.e. when no load is connected)We can solve this using the Voltage Divider Rule.Connecting a resistor to V-out will create a parallel resistor network. We can use the output over sum fo rmula to find the comparable resistance because there are however two resistors.to 1 dpWith this information we can calculate the voltage.V=0.2V 1dp2.19A moving coil meter requires a current of 1 mA to provide full-scale deflection. If the meter coil has a resistance of 100 and is to be used as a milliammeter reading 5 mA full-scale, determine the value of parallel shunt resistor required.REVIEW MEMake the meter useable over 5ma by adding a resistor to switch the range of the meter like you would on a none autorangeing multimeter.This is done by adding a resistor IN PARALLEL with the meter.2.20Two resistors, one of 15 and one of 5 are connected in parallel. If a current of 2 A is applied to the combination, determine the current flowing in each resistor.As there is only two resistors we can use our product over sum equation to find the total value of resistance the parallel network provides.Using this we are now able to find the voltage.Now we can find the current through each br anch,I1I=V/RI1 = 7.5/15I1 = 0.5AI2I=V/RI2 = 7.5/5I 2= 1.5A2.21A switched attenuator comprises five 1 k resistors wired in series across a 5V d.c. supply.If the output voltage is selected by means of a single-pole four-way switch, sketch a circuit and determine the voltage produced for each switch position1K1K5VSwitch1K1KVoutAnswer 1V, 2V, 3V, 4V, 5V2.22With the aid of a diagram, briefly explain in your own words Kirchhoffs second law.In an electronic loop the sum of all the voltages around the circuit victorious polarity into account will equal zero.For example if you where to travel around a circuit following conventional current taking the voltage at each resistance including the battery and added all of those voltages up including negative voltages the sum would equal zero. We would see that the battery would give the circuit charge a potential drop while all of the resistances would dissipate this force.2.23Using Kirchhoffs second law, determine the value of e.m.f. (E) in the circuit below.E+5=14E= 14-5E=9V2.24Using Kirchhoffs laws together with the use of simultaneous equations, determine the current flowing in each branch of the network shown in the circuit below.Here we are presented with essentially two loops of current where readings in the connecting part of the loops will be affected by one another. We will use Kirchoffs laws to solve the problem by first treating the current as two separate loops.We use simultaneous equations to find our two unknowns I1 and I2 .Loop TwoE2 = I2r2 + (I1 + I2)R2 = I2 + 4I1 + 4I22 = 4I1 + 5I2Loop OneE1 = I1r1 + (I1 + I2)R4 = 2I1 + 4I1 + 4I24 = 6I1 + 4I26I1 = 4 4I2Substitute I1 into the second loop.AmpsAs we have obtained I1 we can now work on I24 = 6I1 + 4I2R=I1+I22.25Analyse the circuit shown below and determine the following parametersa) The current in each branch of the circuit.I1 =I2 = 1.233Ab) The voltage across the load resistance.0.426c) The power dissipated by the load resistor.P=d) pulmonary tuberculosis computer software to verify your results.26) A temperature demodulator is connected into a bridge measuring circuit as shown. If the value of the sensor is 110R at 0oC and it increases by 0.2% for every degree the temperature rises and falls a correspondent amount if the temperature drops.What voltage will be output on the voltmeter when the temperature is -(a) 25oC(b) 100oC(c) -40oCBuild the circuit using Multisim and constitute your answer to part (b) is correct.First we will calculate how the changes in temperature will affect the resistance of the sensorNow we must find the voltage for the left give way side of this wheatstone bridge.V1=3VAnd now the right hand side of the bridge, this will vary each time as the resistance of the sensor changes. Firstly we will be doing question a) with the sensor representing 115.1The reading on the voltmeter will be the difference between those two calculationsb)Now we continue the calculations for the second value of resistance for the se nsor.With the sensor representing 132The reading on the voltmeter will be the difference between those two calculationsc)Now we continue the calculations for the third value of resistance for the sensor.With the sensor representing 132The reading on the voltmeter will be the difference between those two calculations2.27 For the Wheatstone Bridge circuit below, what value of R1 will produce a balanced bridge?Using your calculated answer build the circuit in Multisim and demonstrate your answer is correct.2.28 A 1m long resistive wire of uniform cross section is connected to a 6V source as shown.If a sliding contact is placed 0.35m from one end and connected to an unknown e.m.f. then no current is measured on the ammeter.A) What it the value of the unknown e.m.f.?This can be solved using the voltage division rule.