Electricity Notes PDF

Charges Electric There types of charges two are (t) positive : C-) negative neutrons neutral overall > electric field physical field : that surround ( é) } electrons negative * ( no) } positive protons ( pt ) •••o • electrically charged particles charge = pt These exe...

Charges Electric There types of charges two are (t) positive : C-) negative neutrons neutral overall > electric field physical field : that surround ( é) } electrons negative * ( no) } positive protons ( pt ) •••o • electrically charged particles charge = pt These exert . e- = force on + a lines help how visualise a a charge a - found 2ᵗ away from the 1- It = gained / lost be can , exchanged . net : positive charge negative charge points a - charge towards - the charge region µ when there is more than I charge in aresion.fieioii.es curve response to or ions charged particles * charges forming influences surrounding will , The electric field from positive charge points _ nucleus , not shells in + other The electric field from field found in - y other either away towards like charges ( 2 positives / 2 negatives ) electrical conductor : a A ✗ AM in (typically material where electrical charges repel K ) electrons \ opposite charges ( positive % easily transported from are atom to atom as negative ) and have low they to attract K resistivity - e example most metals : and graphite ee- e- éé e- : a electrical material where e- e- example plastics styrofoam , , rubber , glass wood , : • • : an imbalance of charges on or within surface a e- of materials Hiiftini ifdii - t t - + - t . t .Ñj §]y± t + w • . when balloon charges evenly distributed hair and balloon i. e.pt e- = it electrons move , neutral electrically + together + i. is causes an : + : : × - _ - is made up of molecules St H and S polar - O when . will × I 8+1-1 align faces balloon is so charged electrons in balloon flow applied move away . repulsion _ + t + are - + + + I + hi+WW¥ m Am + + + + + + + + - - + ' _ + - i " n - . . I rubbed to balloon hair balloon on as it has hair balloon has × - + + r " / - molecules uncharged t ' + are electrons available to it is able to + × µ is + + " electrical insulator , very mobile ) free + • " × until remain no electrostatic friction from ' applied is covalent bonds and voltage a - × µ water . × - × Ix - + × - P electrostatic attraction t + lost electrons - t - neutral wall electrostatic attraction , , pt pt × , + - net positive charge e- = net negative charge - × m - - + - W charged balloon - t t × + x - - × - I t t , × - = - × , × , e- - + , - t electrons gained + - × ( or very few i. • W + _ lattice , - " + - : - and hair electrons build / typically form most insulators • + " t t I - + voltage a around the metal move highresistivity - + t - - I - - have they r - This imbalance . positive charges not are to e- + 1- . as travel when easily can free are • : 8 s : • : D : • • electricity e- • D B static * easily transported not ⑥ : thus they ee- e- are delocalised valence electrons e- ee- e- electrons) charges (typically ee- e- ee- e- e- e- e- electrical insulator e- e- - - × ✗ I É t w electrostatic attraction t the balloon's repelled the - electrons wall's electrons . ?⃝ ?⃝ Electric Circuits electric circuit conducting path for transmgittin continuous : of convention current current flows from terminal ( cathode) to the electrical fields ? . positive negative (anode ) why ik terminal of current electric of of of of of .☆- electron flow - electrons flow from - 1- Hk terminal (anode) to 1 from positive to ? why are + electrons and \ $ . components Battery of made : circuit : a of up i i + terminal positive wire : which 1 &] battery a metal . 2 ** switch in bundles I . device which 2. 3 or . * light ( non bulb - ) Ohmic : device and II. 1 . wire either it * used to between two points symbol • - , , 3 4 in , tungsten measure a is used interrupted , electrical circuit . heat ) * / as is extra electrons in ' dies ' anode " " on " or flow) no , off " By En r open (A) current . Measured in Ohms (R) High resistance long -_ thin nichrome , short thick copper wire wire , . it has potential : non - causing slower resistance current to than Ohmic through voltage , resistor high melting point very difference volts electrons flow , increases , . in battery and / because it heats up its light Measures , 0×-2 i. e. reduces ¥ increase as battery closed circuit understanding of * low resistance inside is thin glows producing ( V) the circuit to be closed ( continuous flow) r > resistance in volts whether the circuit is controls open a electrons replenish allowing open ( or in produces light ( and provides measured positive terminal no more to device which : material of is flow the reverse by closed As it heats up , instrument 2 which filament MO → oxidised, there V ↑ ↑ a : is Ohmic resistors have constant resistance symbol Voltmeter area , electrons gains : cross-sectional (EMF) atom to atom , attracted to as in using less conductive material increasing length of conductive decreasing electromotive force is its \ resists the flow of electrons by do this can 1. batteries rechargeable insulator to reduce loss end . metal arranged typically wrapped and * the conductive wire and reduced of the metal most once , 1 through move electrolyte gets - symbol * , oxidises and loses electrons symbol / ( Ohmic ) : output can electrons to flow allows symbol resistor * electrons travel toward cathode . * cell electrons . 3. 4 or ( usually copper) conductive metal * chemical energy to electrical energy + ( cathode) * * output ( potential difference ) negative terminal ( anode) - * of cells converts series a The maximum symbol * a they attracted to positive end are pushed out of negative - - MM negatively charged voltage is applied are when negative the positive terminal (cathode) ! propagated negative electrons ) charge ( typically or Ammeter ( V) : instrument an electrical used to circuit current measure . Measures in Amps (A) in symbol + i f- Voltmeters typically on a are connected either side , , of 3 "ˢ☐• 2 , 4 I + connected : in series I battery or resistor \ - : ammeters anywhere Charge current & Current the rate of flow [ or Cs charges (electrons) passing point in of " Q = t which in change conductor of property a a C Charge - an body a experiences electric field force a * or electron As proton ② time (s ) in charge = of : = -1.6×10 +1.6×10 = If there is current of 10 amperes a E- 10A in I= Q Q= ? f. 60s × = C or C or - e e for 10 minutes , how much charge flows through the bulb ? Q= It t 10min = '" elementary charge circuit a - Number of electrons Ne Exampleprobtems 1 '" _ 10A = 600s = 600s 6000C 1min 2 charge of A 6500C passes through circuit a 1- = ? I= Q - 1. OK × = 60min 60s × th A = - Q= 25C 12A is flowing filament E- I= Q 12A circuit a at a point certain I= Q through in 2.2 1.8A ≈ How many minutes . Q t= 25 As = I = = 105mV circuit 2.2×105 mtx I has V e. = 1.6×10 263.168 = to flow does it take 25 coulombs 1min ✗ = past this point 4.4min 608 0.095A t= 5.5min 60s × - = a electric bulb an '" ( 12A)(20s) N= Q 240C How many electrons 2408 = 1.6×10 e of resistance I ZZOV . C V = ISR . What 220 V = R = - '" charge 1.5×10 = passing through a " electrons ¢ flowing through is 5.5 minutes ? in Q= It 14.7A 1552 = 330s = are Q= Ne 1000mV 12=1552 of filament Q= It t ✗ a ? 20s 20s = ✓= in 0.095A cross-section of the A . 3600s t Q= ? 5 flows 1000mA A current of f. 1.8 / Cs -1 = flowing ? t= B°Ñ4 6500C = the current 1min 1 1- = 95mA × Calculate . 3600s = A current of 95 milliamps 3 hour 1.0 t Q= 6500C f. in = ( 14.7A )( 330s) 4851C ≈ 4900C 1min 6 How Q = does it take long V 2500C 2500C a = = 14kt = Q 12=2052 C- C- 1000J × = to pass charge I 14000J = 2500C 14000J = through V a = R 5. GV 5. bv t= 14kt of energy Q = I 205L = if 2052 resistor 2500C = is being supplied ? 8928.6s ≈ 0.28A 8900s or 0.28A = 2. 5h I KX A 250W 7 is being appliance carried in is connected ✗ 1000 V = 200 V 1 KX 35min circuit . The potential P= VI ✓ = 0.20kV = a ✗ 60 s 1min difference across the appliance is 0.20kV . What 35 minutes ? D= 250W 1- in I =P I= Q = V = 2100s 250W 200 V = 1.25A t Q= It = = ( 1.25 A) ( 2100s ) 2625C = 2600C charge ? Ohm 's Law The ~ Charge (C) ✓ in ( J) two ) JC I = " ) resistance R • the ~ amount the flow current points of r of opposition to electric current (E) passing point a in Cs conductor a R =p or charges " A subwoofer needs -1=5500 mA/ 1 × R= ? An A V R= 5.5A = 1000mA = length of wire A area by passing mv ) needs to be ✓= ? 4 I. = R = , through coiled a 1100 = 1000mV × I to the current happens t.SU V. ? I. 9KC charge in 352 circuit a IR = V = 1. SV = flowing through is 1000C 0.5A = 3h R Q= 9K¢ ✗ circuit a wire R= 850mA t= 2min If 1200J = 0.8552 1200J I = 7.5×107 # is supplied to V. = 3V V. 1 A I 3500ms × s I, = ? I,= V P= 1214W ✗ = R, = 652 1000W = = = 3.5s . copper = a) How would , what 3V a battery resistance and V 0.5A = no 2 minutes in . doubled ? A change ! Calculate the is voltage . (75-1×0.8552) GOV ≈ 63.75N = is 262.5C = 263C is supplied by 110 V. I =P 12000W = same their resistance length . for 3500ms = calculate the resistance . R= V I 1200J 263C is , V=IR Q 4. 6V = 4. 6V 0.06152 = 75A its resistance ? R= 12000W 110 V V have the What flowing ✓ = C- (75A)( 3.5s ) = = wires 1152 is wire ✓= IR = IIOV Two with GR R 7- 75A 1 kW 7- resistance 106mV 3 = 1000ms A 12kW appliance the IR = Q= It × ✗ replaced is circuit where 7.5×10 µ -1 a 106µA ✓ its 75A = of energy = = Calculate 120s 120s = 1. I battery t.SU a 9000C = = if t I R × C- t= . ✗ I= Q 9000C = 1000m¢ 6 If the resistance of . with 850m52 of resistance 1k¢ 5 its coil through to it ? (100 A) ( IIR) = 352 A 5500mA Rm 20N = 12=1.852 = m2 5.5 A 100A current a applied I. = 100A V. of current ✓= TR What m resistivity a / IOV I electric heater works voltage ( in 3 ' ✓= IR ✓ = HOV 2 of 110 V to push voltage household a VA l Exampleprobtems 1 or - the rate of flow of Potential Difference aka constant with Resistance circuit a the current in or potential energy between conductor is a V the amount of " Q energy across Voltage C- = voltage Vy = - 110 V = 1. OR 109A 109A One wire has double the diameter . b) how differ ? would the flowing through current differ ? them I wire 0=2 1- = Wire × × ,Trz =3 / m2 . R=p l 3 0=4 A = = R =p ITRZ 12.6m l 12 ' . : thinner has more ~ wire 4× resistance wire × I= ✓ R wire = V ' 13 I= It y ✓ R = V ' 112 . : wider have ~ current R wire will 411 more Series Series circuits only : path for / b Parallel Circuits Parallel circuits.mu/tip1e,paths(branches-forur entt,of1ow current to flow c 2 - R, 1° a - - 2 I ! 1 vs ! , 3 R + + Rz h ' 1 I C 4 4 3 2 I VoHag# 3 - g the 12 of PD sum equal f } to the PD provided by V = IZV = , ex : V2 GV 1- V3 1- V4 + 4 V + ZV VoHag# is b ex : f n o PD across each component = IZV = V2 V3 = IZV = IZV = = V4 Resistance ( R) g Resistance Total sum resistance of each of all components R, = Rz + t I one R} = 10h ' = I ex : = 2 4h I I + R, RT y 652=352+252+152 ex : more branches same voltage I ' any single resistor the is GR RT equal the combined resistance of resistors is less than that of ( R) is IZV o c > voltage V , e less resistors more 4 power source a 12 I } 1 component each across R 3 2 - Rz R2 I I + 1 + branches R} : : :: : : : i. . - I + 6 6 More lower resistance vs 6 1 currenl.CA#- current.CA# Current at each A, = 2A ex : = point Az = As 2A = 2A in the circuit equal is the current from the larger is A, 6A ex : than the current = Az + A} = 2A + 2A + + : :: source A- 4 in a More : - . . . . . branch branches 2A Higher current Advantages ✓ as current is components i. e. ✓ current are ✓ bulbs all equal everywhere behave same increases , wiring is ✗ as only path if 1 , fault is a brightness is disconnected if more cells I switch controls all , entire ✗ cells don't last ✗ if more resistors resistance ✓ there circuit ✓ as are increases , long if there only ✓ added dropping ✓ ✓ simple I is a fault Disadvantages in a branch , that branch is disconnected longer cells last in current components ✓ , similarly , added simple Advantages Disadvantages parallel as is constant multiple switches can independently control specific branches /components each component easy to multiple wires × voltage cannot be increased when connected voltage relieves same edit (add/remove voltage components) required ✗ by × as adding cells increasing increase parallel in load does not resistance become overloaded , , circuit leading can to fire Series Parallel Circuits vs Exampleprobtems Calculate I a) current and b) voltage the 7- V | RT a) " " " Rat Rb = a b ✓ I= 40052 7- V i. , ,, , go , 4V = (0.01-1)/20052 ) = ZV = 0.01A = 70052 R , Vc= IR (0.01-1)/40052 ) = IV = = Vb= IR (0.01-1)/10052 ) = 70052 = I below b) Va= IR Rc + 10052+40052+20052 = look resistor in the circuit each across , a, ,, , a ,, ,,, , ,, C , the a) Calculate 2 current b) current and ✗ I a) 7- V 1 40052 Ia 20052 lb = ✗ Calculate resistance X 10052 20052 la lb / - I The 100h battery is - , set to ION both 3. OV . Calculate Ry = ZOR ammeter reads and 2A closed b I , = 0.018A A = 7- A 0.035A = 200 the 1 = 20052 X 20052 = I = + R, I I = R2 I + 20052=66.71 = 3 10052 20052 50052 series + Rtparallel I = 50052+66 > R = 566.752 = ✓ R . potential difference both resistors across a) a lb I=✓_ Rat Rb = 102+202 = 305L the . a) Ra Va = = = 30 = GV = = 0.0 / A 566.752 . Calculate the voltage change ? 9V Vy = Va 2A c) does = Vb = 9V the 4.552 i. change 4.51 ( IA) ( IOR) = = /OV = 2 equal bulbs of the b) switch closed I I = R 9V bulbs ZOV the switch when is change ? c) switch open P= = = VI ( 9V)( 2A ) 18W 4.552 resistance does not ( IA) (ZOR) = brightness = = Vb=IRb Va=IRa Rba -1=4.552 Va 9V /A and resistance of switch closed I = = 305h R of the ammeter value switch open VT ; - 7400 V = ✗ Rtparaliel 10052+40052 RT = switch is open when the b) does . za RT 12 ' open = I a 9V 1 l The = 200kt " " 6 1 Rat Rb = series 1 5 0.07A circuit below in the RT 10052 2005510052 5052 X § I + 1 _ b GV : Ic 0.12A I - a V = R Rb 20052 = - = I + I + = Calculate the current of the ammeter = Ib 400 Ra X I 4005L I + I = 5052 12 -1=5052 1005h 49 A = 400752 Rx I - 4 = 100 R TV = = RT ✗ A - l 1 ✓ R 1 ; 7- 7 I i = R 400 R = - 3 V = I + 4005220052 10052 Ic Rc I + Ia b) I + Rb I resistor in the circuit each I + Ra = 10052 ; I = RT ✗ I flowing through = 2A switch closed P= Vb = OV = Rb = OR = VI ( 9V)( 2A ) 18W ' . . Same Power P Power (J ) energy used energy € = t dissipated transformed or P or ( Js 1) by (E) passing points 120J uses C- = 120J t= D= C- = = Your house ✓=/ 20A I I = F- 1500W 3 A motor has a A = of 12kV supplies = = 110000m¢ × I 0.2 KJ No current is the flowing through ? 0.15A = V of current . . 12.5A will blow the circuit . and current a which , pushes a 1.2A 110000 mc Calculate the . charge in motor power of the circuit a I= Q ✓ =P t t I 200J = 110C = . 180s t.TW = 0.61 = over its 3min Calculate . = 1 1.82K , t.TW = 180s 180s = of P= C- 110C = it across of energy 200J C 60 S × ) 14400W 1000m¢ E- 3min " ( 12000411.2A) = 0.214×10005 ) JC 80W , what is " P= VI 1kt Q it Cs conductor a or two Potential Difference across in 120 ✓ KX source 80 Explain 12.5A = 12000 V = house ? your = power C- in 1.2A = 121 aka between or charges 12W potential difference 1000 V / 4 V 1500W = ✓ ✓ = 12kt × I 10s = circuit a difference potential I =P hairdryer 1500W a in potential energy standard ROV and has fuses ( circuit breakers ) that blow at 10A relieves run you If the . 120J t 10s Can 10 seconds over ✓ = 80V 2 V the amount of A device point a Voltage time Exampleprobtems 1 _ the rate of flow of - ( s) time I • = Cloud ) the resistance over Current W voltage in KV kV 1000 it 0.61A A = 1.8×10 -3kV 1min 5 Your oven uses at 2500 V current of a f- 2500N = 90min 1h × = ( 2A)( 25004 = 5000 # 1. 5h = . . . KIWI 1 kW 1.5K × what will it cost to × 514W use the 112.5¥ = 5kW = 1000W Calculate the power of the 1. SV ¥15 perk,Wh charged are 15 ¥ 60min 6 If you . F- IV I = 2A f. 2A light bulb in the circuit below ? 1. SV . . VT . = 1. SV I 1. SV + = 3V = R =3 V 3h V = P= IV /A 352 = ( / A) ( 3V) 3W = # 7 Calculate the - , power supplied by R ,- IZV 3h A circuit contains - , IZV battery two and 3ohm bulbs in series . I 3h 8 a a = 352 = 6h IZV + I 3h = V IZV = R battery two and = D= IV 2A GR 3ohm bulbs = in parallel . ( 2 A) ( IZU) = 24W Calculate its power . I 1W 3h 352 I RT = I 35h + I 3h = 1.552 I = V R = IZV 1.5h = 8A P= IV = ( 8A)( IZV) = 96W oven for 90min ? .