NEET Booster Week - Current Electricity

Hinzugefügt: 2026-05-17

[00:04:09]What are you shooting for?

[00:04:25]Ready sir.

[00:04:34]>> Hello students. Good evening all of you.

[00:04:38]First of all I would like to all the best for your neat June 21st examination.

[00:04:50]Today in this evening session we are going to deal with current electricity most important chapter in physics and high weightage chapter in physics.

[00:05:04]Current electricity usually two to four questions you can expect in neat examination.

[00:05:15]If you go through the previous neat examination papers, most of the cases three or four questions they are giving in from current electricity.

[00:05:29]In fact on May 3rd recent neat exam they have given four questions from current electricity means 16 questions sorry 16 marks four questions they have given 16 marks and this currentity is so important because of two reasons. One is because of this high weightage and second is this chapter is linked with some other chapters like uh moving charge and magnetic field electromagnetic induction AC circuit and semiconductors.

[00:06:16]This chapter is linked with them also.

[00:06:19]Hence in this current alexity strong foundation is essential dear students.

[00:06:28]And now I'm going to give you some important tips, tricks to solve the problems, elimination methods, timesaving techniques of solving the numerical problems today evening. Now, now we are going to solve together we are going to solve some important needle level problems from current electricity.

[00:07:03]Starting from basic two tricky rank improving questions we are going to solve to be ready with your notebook and you try to solve every problem before I solve it and by the end of this chapter you should feel confident to attack any problem on currentity in the examination.

[00:07:32]So let us play with current electricity and enjoy without getting shock.

[00:07:41]So let us start now.

[00:07:45]In this chapter you need to focus on W's law drift velocity.

[00:08:03]Oh law, drift to velocity, series and parallel combinations of resistors.

[00:08:15]These are areas you should focus more series and parall combination of resistors.

[00:08:25]Kitchop's laws won bridge meter bridge and electrical power these areas you have to focus and if your concepts are clear on these areas then you can solve any problem on current electricality quickly.

[00:08:53]So now let us start solving the problems through problems. Let us discuss the concepts and the tricks and tips.

[00:09:04]The first problem if you look at this circuit has given a junction of three resistors and is asking in the following network find the potential at the junction W.

[00:09:22]And when you look at these three resistors junction is showing that indicates that this problem is based on kits first law. What is that first law?

[00:09:37]Law of junctions law of currents. It is based on kits first law. Law of junctions law of currents. The total current flowing into the junction is equal to the total current flowing out of the junction.

[00:09:56]This law is nothing but law of conservation of charge.

[00:10:02]And you know the second law kit's second law of voltages or law of loops that is law of conservation of energy. First law is law of conservation of charge.

[00:10:20]So in this problem when you look at that junction immediately you'll get idea that we need to apply Kitop's junction law. Kitop's law along with Ohm's law we should apply to get solution for this. So let us assume the potential at this junction is V K.

[00:10:50]And let us assume the current flowing from A to W is some I and this current here splits into these two branches.

[00:11:02]And let us assume the current from O to B as I1 and the current from O to C as I2.

[00:11:13]Then according to Kov's law, I is equal to I1 + I2. I can write. This is Kov's junction law. Now comes Ohms law.

[00:11:29]Current flowing between A and O must be equal to the potential difference between A and O divided by resistance between A and O. That is Ohm's law.

[00:11:45]Potential difference between A and O.

[00:11:49]Write it. Potential at A is 8 minus potential at O is V. We assumed 8 - V by resistance between these two points is 2.

[00:12:05]Similarly, write I1.

[00:12:09]It is the current flowing between O and B. Potential difference between O and B.

[00:12:16]Potential at W is V. Potential at B given as 4 volts. there V -4 by resistance 4 then write I2 potential difference between O and C V - 2 by resistance between O and C is 2 substitute this here 8 - V by 2 is equal to V - 4 by 4 + V - 2 by 2.

[00:12:58]Okay, you can cancel 2. Now here 2 will cancel here. I am writing 8 - v = v - 4 by 2 2 * cancel here + v - 2 then 8 - v don't take lcm here it'll become lengy 8 - v equal to this is split into two fractions v by 2 - 4 by 2 4x 2 that is 2 + v - 2 then v 8 - v is v by 2 + v 3 v by 2 - 2 - 2 adding -4 you'll get then bring Four here v there you will get 8 + 4 12 = 3 v by 2 this minus v not bring here + v it is then you'll get 12 = now take lcm 5 v by 2 that implies V is 24 by 5 4.8 rate.

[00:14:42]Fourth option you are getting. Now in this question instead of asking the potential at the junction W they may ask any current in any one branch here.

[00:14:59]They may ask find the current in this 2 ohm resistance or find current in that 4 ohm resistance. anything using this you can find. Now substitute V not value in any one of these equations that corresponding current you will get suppose if they ask find the current in 4 ohm resistor in this network. Now this I1 in this V value you substitute and find the current V 4.8 8 - 8 by 4.8 by 4.2 ampere you'll get 4.8 - 4 by 42 ampere current flowing in this like that you can find even current in any one branch in this.

[00:15:55]Okay.

[00:16:15]Now let us see the next problem.

[00:16:19]See again Kit's law, second law, loops are there. Now here previous problem junction is there. Kitch's first law, junction law we applied. Now if you look at this network loops are there. Hit's loop law voltages we need to apply. Now what is asking in this network?

[00:16:49]The cells have zero internal resistance given there.

[00:16:56]The currents passing through resistances R1 and R2 R is asking current in this R1 and in this R2 is asking.

[00:17:10]Then how to solve this? You take current some current from this battery. Current direction is always shown from positive terminal of battery. No. So here you take some current I1 coming from this first cell.

[00:17:30]If current coming from pass two terminal of cell if you take it as I1 the current going into the negative terminal of the cell must be I1 only. No same I1 should go into its negative terminal.

[00:17:46]Similarly from this cell positive terminal current coming out I'm taking as I2.

[00:17:56]Then the current going into this negative terminal of the second cell must be I2.

[00:18:04]Now this I1 and I2 currents combine and the current flowing in this middle branch will be I1 + I2.

[00:18:16]This I1 + I2 at this junction again will split into two branches. This side I1 should only come I1. This side I2 should be there.

[00:18:35]So first like this you find out how the currents are flowing in this network first then apply kit's loop law.

[00:18:51]Let us take first this loop. Let us name this A B C D E F.

[00:19:04]Now for this closed loop A B C D F I'm applying Kshop's law A B C D 8 A B C D A per this closed loop kit of second law I'm applying. So let us start you know remember the sign convention used here for any resistor.

[00:19:38]When you are writing the potential difference across any resistor, what is that? Product of current and resistance. No potential difference.

[00:19:52]For this the sign convention, if you are going along the same direction of current in that conductor, then this potential difference is taken with negative sign.

[00:20:08]If you are going opposite to current then this potential reference I into R taken as positive sign.

[00:20:20]Whereas emf of cell for that when you write if you are going from positive to negative terminal of the cell it's emf must be written with negative sign positive to negative terminal side if you are going take its emf with negative sign.

[00:20:47]If you comes from negative to positive terminal side of cell the sign convention is take its emf with positive sign. So I'm applying this start from A go to B then to C then to D and come back to A travel along this and write the algebraic sum of emfs and potential differences and equate that with zero.

[00:21:18]So let us start from A. Start from A.

[00:21:23]Move up to B. Between A and B, what is there? What is there here? Between A and B, resistor is there. That means there is potential drop between A and B.

[00:21:39]Potential difference is there. That potential difference is I into R. And we are going in the same direction of this current A to B we are going means in the same direction of current we are going.

[00:21:56]So with negative sign you have to write it. So minus I1 + I2 into 20.

[00:22:07]Now travel from B to C. Between B and C there is no resistor here and there is no cell also. That means there is no potential difference between B and C. B and C are at same potential.

[00:22:23]Similarly C and D also they are at same potential C and D also.

[00:22:34]Then move from D to A. Between D and A, there is a cell and that cell doesn't have any internal resistance. C gave it zero internal resistance. If it has internal resistance some small R, you have to write potential difference I1 into R. You have to write I1 into R in the same direction of current. So if you we are going now minus I1 R we have to right but there is no internal resistance for this only emf and while going from D to A you are going from negative to positive terminal side of that cell. So it emf must be taken with positive sign + 10.

[00:23:23]This is the algebraic sum of emfs and potential references in this loop.

[00:23:33]This value must be equal to zero according to kit's second law which is nothing but law of conservation of energy.

[00:23:50]this now - 20 I1 - 20 I2 - 20 I1 - 20 I2 + 10 + 10 I'm taking that side is - 10 you are getting then what you'll get - - cancel 0 0 cancel you will get 2 I1 + 2 I2 equal to 1 one equation you got now Kov's law second law loop Loop law you apply for second loop A B E F A B E F A for this loop you apply.

[00:25:03]Similarly start from A go to B then to E then to E F and come back to A.

[00:25:12]A to B minus I1 + I2 into 20 then B2 E there is no cell there is no resistor B and E are at same potential and E to F there is a resistor 20 ohm is there so it's potential difference you have to write in the same direction of current we are coming so minus - 20 I2 - 20 I2 then move from F_sub_2 A F_sub_2 A F_sub_2 A when you are coming you are coming from negative to positive terminal side positive MF you have to write + 10 it's zero So -20 I1 - 20 I2 again - 20 I2 means you'll get - 40 I2 is take that side - 10 - - cancel 10 also cancel so you'll get 2 I1 + 4 I2 is 1.

[00:26:48]This is second equation.

[00:26:52]Now solve these two equations and find I1 and I2.

[00:26:58]Solve these two equations. One equation.

[00:27:01]Second equation. First I'll write 2 I1 + 4 I2 is 1.

[00:27:09]First equation 2 I1 + 2 I2 is 1.

[00:27:17]If I subtract these two 2 I1 2 I1 cancel 4 I2 - 2 I2 is 2 I2 is 0. I2 you got I now this in first equation I to0 you substitute in first equation I to0 then according to first equation from first equation what you are getting 2 I1 is equal to I20 now this is zero 2 I1 equal to 1 you are getting that means Means I1 is 1 by 2.5 aere I1 I2 zero. Now the question is what are the currents passing through resistances R1 and R2?

[00:28:13]In R1 I1 + I2 is current I1.5 I2 is 0.5 + 0.5 current in R2 is I2 that is 0. So 0.5 second third option.

[00:28:41]So Kitschov's first law first problem Kitab's first law application and the second problem is Kitop's second law application kit's law very important you need to practice you need to practice more this chapter in fact physics means you need more practice is if you have more practice in physics then you can quickly understand what is problem and you can quickly solve and save time. Practice very important for physics.

[00:29:26]Practice makes all perfect.

[00:29:33]Okay, I'm going to next problem. Third problem.

[00:29:46]Third problem. A wire of resistance R is bent into a circle of radius small R. What is the equivalent resistance between those points X and Y?

[00:30:03]That is asked.

[00:30:06]So there is a straight wire of total resistance R.

[00:30:14]This straight wire of total resistance R is made into the form of a circular loop means this complete circular loop resistances are.

[00:30:32]Now what you're doing you're finding the effect two resistance between these points X and Y means what is the meaning of this between X and Y if you connect a battery then what is the effective resistance of this network is asking that see if you connect a battery like this between X and Y the current coming from battery at this point I splits no into two branches like this. I2 I1 like this.

[00:31:11]I2 will flow like this. I1 like this and then flow into this.

[00:31:16]I1 will come to this point Y. I2 will flow like this and come to this point Y.

[00:31:23]Again at the point Y, I1 I2 combine to form I.

[00:31:30]again I and this segment X W this W X W Y this segment X Y this segment both are under same potential difference they are in parallel combination so if this resistance of XWY if it is R1 one X Z Y resistance if it is R2 these two are in parallel combination no so what is the effect to resistance between these points X and Y R1 R2 parallel so R1 R2 by R1 + R2 now effect two resistance when two are in parallel but R1 plus R2 means this total loop resistance R only. So it is R1 R2 by R.

[00:32:41]What is R1? What is R2?

[00:32:47]See this resistance is row L by A where L is length of the resistor and A is the area of cross-section. Row is resistivity.

[00:33:02]These two segments are made of same material.

[00:33:06]So row same now area of cross-section also same. Now both are same area of cross-section not circular area that wide area of cross-section same now for both that means resistance proportional to length of the segments.

[00:33:28]But what is the length of arc in a circle?

[00:33:32]You know that length of the arc in a circle formula you know now r theta means length of the arc is proportional to angle.

[00:33:46]So distance proportional to l proportional to theta like these tricks you have to use should not make lengy calculations r proportional to l proportional to theta.

[00:34:01]So r1 is to r2 must be equal to theta_1 is to theta_2 means r1 is to r2 is theta_1 some alpha given there alpha alpha alpha this angle mean this back side remaining angle 360 minus alpha 2<unk> radians minus alpha.

[00:34:33]So theta_1 is alpha, theta 2 is 2<unk> - alpha.

[00:34:44]That means this total resistance R is made into two parts R1 R2 according to this ratio.

[00:34:54]So R1 is how much?

[00:34:58]Suppose if the ratio R1 is to R2 is 2 is to 3. Suppose if you have like this what is R1? Two parts R1 2 out of five parts 2x five of total resistance 2 by 2 + 3 5 into total resistance like that R2 three parts R2 getting 3 by 2 + 3 pi into total resistance like that you do this R1 is alpha by alpha + 2<unk>i minus - alpha alpha alpha total angle 2 pi alpha by 2 pi into total resistance what is r2 r2 it is 2<unk>i - alpha by sum of these two means total angle 2 pi into r so this to substitute here if I substitute here R1 into R2 when I do this in the denominator 2 pi into 2 pi 4² I'm getting ah see here see here here only in one option 4 pi square is there in no other option only in one option is there first option so this is answer elimination method you apply first option is right substitute r1 r2 these two here and simplify get first option That is this problem.

[00:37:23]finish.

[00:37:25]Now I want to discuss some more points here.

[00:37:29]Resistance we are discussing now why bent into the form of circle.

[00:37:38]See here they can ask you here some more things stretching stretching the wire.

[00:37:56]A wire is a wire of resistance or is stretched.

[00:38:08]Stretch should stretch.

[00:38:22]If you stretch wire, very important concept this is if you stretch this wire its length will increase but at the same time its area of cross-section decreases.

[00:38:38]Two changes are there. Stretching two changes are there. Length changing length increasing radius decreasing two changes. But in question what they do a wire is stretched so that its length is increased by 20%. A wire is stretched so that its length is increased by 4%. What is the new resistance? Only length change they will give but radius also changing. No this problem some students what they do length change they gave length change.

[00:39:17]So R proportional to L resistance proportional to L they will use this resistance is proportional to L under one condition that is when the resistivity and area are constants then resistance proportional to L suppose if I have two separate wides of same area of cross-section Same area of crosssection two separate wires I have but one is long wire one is long other is short length different but area same material is same material also same means resistivity is same row row a they are same length different in this case I can write proportional to L R1 is to R2 is L1 is to L2 because length uh sorry resistivity and area of cross-section same in this case but the stretching different wire same wire stretched length increased area decreases area also changing then area also changing how can you write R proportional to L R1 by R2 = L1 by L2 that's not correct no Area change also you have to consider. No. So how to do this problem? When you stretch wire one parameter remains constant for that wire that is volume volume constant.

[00:41:07]So resistance royal by A if you take suppose you want resistance length relation when wire stretched then what you do is this you multiply with L and divide with L L by L you write into area into length volume so you'll get ro² by volume now in the stretching process in your row same volume same. So resistance is proportional to square of length you have to use when Y is stretched are proportional to L² you have to use similarly Y stretch it's radius becoming half what is the new resistance they are asking Now radius change they are giving length change not given then what you do r is royal by a length I have to eliminate because they have not given a length they have given radius change you know not length change I have to eliminate length in place of length volume I should bring there so what you do is multiply this with area of cross-section and divide with area of cross-section Then this will become row into length into area of cross-section is volume by a² row same v same now in stretching process are inversely proportional to area of cross-section squared area of cross-section again<unk> r²ared now for a while p<unk> r² if I write finally what I get resistance inversely proportional to r power four you will get. So they will ask questions based on this resistances wire bent into loop circle it stretched what is the new resistance suppose they gave resistance length length a wire is stretched so that length increased by 4% 6% like that small percentage change given in that case you can write like this now delta R by R into 100 is 2 * of delta L by L into 100 and you can solve it.

[00:43:51]Similarly for this r proportional to r^ -4 delta r by r into 100 is -4 into delta small r by r into 100 for small changes only.

[00:44:15]Suppose if they give length increased by 20% what is the percentage increase in resistance 20% change in length they are giving now don't apply this now what to So R proportional to L² when it is stretched you write this R2 by R1 is L2 by L1² you write you take initial values 100 R1 100 you take here L1 100 you take 100 100 you write initial values when you solve percentage problems and length increased increased is increased by 20% given so what is the new length L to 100 + 20 120 all² and when length increases resistance also increase no R 100 I took now R200 plus some X you write and find X this is procedure when the percentage changes more when the percentage change is small Use this delta r into 100 = 2 into delta l by l into 100 for stretching process I'm talking from this fine x value fine fine f find you'll get answer for Okay. So, I'm going to next problem.

[00:46:22]Next problem. Next problem. A wire of 10 m long has a resistance of 40 ohm. It is connected in series with another resistance box of unknown resistance capital R and a tool to sell the potential gradient.

[00:46:44]Oh, potential gradient they are giving.

[00:46:52]See, see this students potential gradient. What is potential gradient?

[00:46:58]Potential difference per unit length. It is its value given there.1 uh units another point in this current electricity always current micro ampere milliampere mill to per cm like that only mostly they'll give like this only units also you need to be very careful while doing problems in this chapter you need to observe that mill all these have to convert them this potential gradient event. Now see this problem is one resistor is there.

[00:47:41]This is connected to another unknown resistor R in series and this combination is in series with a cell 2 to cell and that cell doesn't have any internal resistance there. No internal resistance.

[00:48:03]He is giving potential gradient of this.

[00:48:08]Its resistance also given 40 ohm. This is and its total length given 10 m long. It is it's potential gradient given this value.1 mill 10^ -3 per cm 10^ -2 mh simplify this simplify this I think you'll get 10^ -2 to volt per meter that is potential gradient. I want total potential.

[00:48:58]See when two resistors are in series with a cell that uh two resistors will share that potential reference V_sub_1 V_sub_2 potential divider rule you have to use current same in series current same current will be there in both V = I constant here I same in both means V proportional to R for them.

[00:49:33]Potential difference proportional to resistance.

[00:49:38]V_UB_1 is to V_sub_2 is R1 is to R2.

[00:49:43]V_sub_1 by V_sub_2 is R1 by R2.

[00:49:50]We need to use this when you look at the two restorers in series. one restor potential given immediately should get this idea potential divider rule we have to apply so v_sub_1 let us find v_sub_1 v_sub_1 because this positive wire potential gradient given now v by l is 10^ -2 vol per m find potential difference this 40 ohm potential difference you find Right.

[00:50:25]10^ -2 volt per m into length in meter.

[00:50:31]You're right m only 10 m.

[00:50:36]So you're getting 10^ -2 into 10.1 volt.1 volt. You are getting V_sub_1 V1.1 you got V1 V1 + V2 should be how much how much this V_sub_1 plus V_sub_2 here this battery doesn't have internal resistance zero it is when internal resistance zero internal potential difference is equal to emf only emf of battery.

[00:51:21]So v_sub_1 + v_sub_2 total value must be 2 only.

[00:51:27]Then v_sub_2 is 2 - vub1 2 -.1 that is 1.9 vol.

[00:51:41]So I'm substituting in this.1 by 1.9 = R1 by R2 R1 40 given R2 unknown R from that find R 1 by 19 is 40 by R R is find that R Value 40 into 19 760.

[00:52:44]Okay, I'm going to next problem.

[00:52:53]Next problem.

[00:53:00]Homes law.

[00:53:02]This problem is homes law.

[00:53:09]Graph given now potential voltage current graph given is based on Ohm's law.

[00:53:20]What is Oh law equation v = i or v i graph must be straight line through origin v by i = r and a very easy problem. This is suppose if this angle is theta slope of this graph tan theta must be.

[00:53:55]This V by I V by I nothing but resistance only.

[00:54:11]V by I find that V 4 volt per that current 4 4x4 1 ohm very easy one ohm Physics very easy concept if you are clear with your concept physics easy.

[00:54:58]Uh this is drift to velocity concept.

[00:55:06]A current of 10 ampere is maintained in a conductor of cross-section cm²ared given observe it.

[00:55:17]And the number density of free electrons is 9 into 10 4 28 m power -3 here meter there cm observe that drift velocity.

[00:55:36]What is this drift velocity formula?

[00:55:41]I by N A E in this what is N free electron density means total number of electrons by volume of that conductor number of electrons per unit volume that is that only given here directly In formula based this is current 10 ampere given 10 ampere by 3 electron density 9 into 10 ^ of 28 si unit only there area of cross-section 1 cm² g this we need to convert into m²ared. How much? 10^ - 2 m² - 4 into 1.6 into 10^ -9 kum electron charge.

[00:57:00]This will give you answer H.

[00:57:06]So this uh 9 into 1.6 6.

[00:57:10]Okay. This 1.6 1.6 16 by 10 you write 16 by 10 10 goes up now.

[00:57:27]So what you'll get is this 10^ all let us keep aside 9 into 1.6 9 into 1.6 14.4 four all other 10^ 10 power that I'll separate it six answer should start with six because 14 into six nearly coming 14.4 4 into 6 nearly coming to this one.

[00:58:03]So answer must start with six. It should be more than six.

[00:58:09]First option I think if you simplify this first option elimination methods to apply and save time.

[00:58:38]Okay. Simplification. Simplification you have to do.

[00:58:43]You need a practice. Not me.

[00:58:52]This very important problem. I like this problem so much. This very important problem.

[00:59:01]unbalanced Wstone bridge. This is what is asking here is giving one network and asking potential difference VA minus VB. How much?

[00:59:19]What is this? See this battery P some Q some R some S A B C D four resistors are connected to form four junctions four junctions. See where battery connected? Battery connected between two opposite junctions either AC or BD any one pair of opposite junctions between them battery connected.

[01:00:17]This is vision bridge.

[01:00:22]Sometimes they'll deceive you in this vision bridge. They'll give four resistors like this. Four they'll give and between two adjacent junctions. So they will show battery like this. Uh P by= R satisfied this V sh if you do like that that is wrong.

[01:00:45]This is not that. See here this is different.

[01:00:50]Von bridge means four resistors connected to form four junctions and between one pair of opposite junctions battery connected.

[01:01:07]This is not based on bridge.

[01:01:12]And what is this circuit principle?

[01:01:17]If the ratio of resistances in this series arm P by Q is equal to ratio of resistances in this series of other series ohm RBS is satisfied if the values of these four resistances are such that P by Q is equal to R by S.

[01:01:55]Then this point B, this point T are at same potential.

[01:02:05]B and D are at same potential.

[01:02:14]The potential difference between these two other pair of opposite junctions will be zero. You connected battery between one pair of opposite junctions.

[01:02:24]Now if this ratio satisfied potential difference between other pair of opposite junctions will be zero. These two are at same potential.

[01:02:38]Even if you connect here one resistor in this middle resistor current doesn't flow that is w principle then this uh bridge circuit is set to be balanced w bridge look at this this 4 ohm 2 ohm 2 ohm 4 ohm P by Q is not equal to RS that is unbalanced Vstone bridge.

[01:03:18]Unbalanced Wstone bridge means the potential difference between this pair A and B is not zero. They are not at same potential A and B. There is a potential difference between A and B is asking that. is asking that only.

[01:03:40]So how to solve this very important problem? Practice this current from battery.

[01:03:57]I you take at this joint it will split into two.

[01:04:05]Now I1 I2 find I1 I2.

[01:04:14]You take this point as P.

[01:04:19]You find potential difference between between between P and A.

[01:04:31]That means VP minus VA.

[01:04:36]Find this VP minus VA.

[01:04:40]What is potential difference between two points? Product of current flowing between those two points and resistance between those two points. Current flowing is I1 and resistance between P and A is four.

[01:04:59]I1 into 4. Find this.

[01:05:03]Then find potential difference between P and B. P and B. Between these two you find VP - VP.

[01:05:23]Same current flowing from P to B I2 into resistance is 2 ohm here. But what do you want? VA minus VB means this second equation minus first equation if you write you'll get answer VA minus VB VP - VB minus of VP - VA is nothing but VA minus VB do this. So first I1 I2 we need to find I1 I2 first that I should find.

[01:06:13]How to find I emf by resistance in the external circuit plus internal resistance.

[01:06:25]Many times I told you we used this earlier.

[01:06:33]What is emf 10 volt by resistance in the external circuit? This four and two in series. Now same current is there in them. 6 4 + 2 2 + 4. These two is in series again. Six. Six and six are in parallel.

[01:07:00]Six. Six. These two are in parallel.

[01:07:05]Then what is the FA2 resistance of this?

[01:07:11]6 into 6 by 6 + 6 R1 R2 by R1 + R2.

[01:07:20]I know you know this very well. Three you will get now.

[01:07:29]Plus internal resistance uh uh print mistake internal resistance here coming as 1 ohm. You take it as 2 ohm 2 ohm. You take 3 + 2 2 you take internal resistance then 10 by 5 2 ampere is the total current.

[01:08:01]Now here and here what are the resistance total in this in this branch E4 + 2 6 in this branch also six and what is happening here 2 ampere current is coming into this branch where there are two resistors equal resistors When resistances are equal, this current will split as I1 I2 equally. Now because current inversely proportional to resistance when potential difference I inversely proportional to R resistance is equal. So current will split equally.

[01:08:52]So in this 1 ampere in this also 1 ampere will be there 1 1.

[01:09:00]So now what is this I1 1 ampere what is this I2 1 ampere then what is potential difference between P and A 1 into 4 4 vol what is potential difference between P and B I to 1 1 into 2 2 volt now V A - VB equal to VP minus VB VP minus VB 2 old you got two you got VP minus VA four you got -4 so answer is -2 -2 fourth option finish going to next.

[01:10:14]Yes, this is very important.

[01:10:20]Almost all models till now we covered.

[01:10:22]Now this is electric power. Last model important models.

[01:10:35]It takes 12 minutes to boil one liter of water in a electric kettle.

[01:10:43]Due to some defect, it becomes necessary to remove 20% of the turns of that heating coil in the kettle.

[01:10:56]After that repair, how much time will it take to boil same amount of water? One liter here also one liter only. How much time you asking? Power, electric power, heat produced. What is electric power?

[01:11:15]V² by R.

[01:11:17]I² R.

[01:11:26]V I these are the relations.

[01:11:29]Total heat produced.

[01:11:32]V² T by R I² R T V I T power is heat produced per unit time.

[01:11:48]Heat produced per unit time.

[01:11:52]Now in this question if you observe which formula to choose here V² by R I² which one that depends on out of V and I V and I which is constant you observe in both cases.

[01:12:22]Then that formula if you use easy it is one coil you are connecting to mains then 20% of that coil you are cutting and removing the remaining coil again connecting to same means means what is same voltage V is same What is changing here?

[01:12:50]Resistance changing that is this problem.

[01:12:58]So power equal to V² by R heat = V² T by R. This you have to use now.

[01:13:12]Now to boil water what is the amount of heat required to boil water in calor cal calorimetry we will do now ml where l is latent heat of water m is mass of water this heat is supplied by that heating coil v² t by r so M L = V² T by R.

[01:13:50]Now what is asking us to find time? Time time. Let us make formula for time here from this cross multiply M L R by V².

[01:14:07]Now in this problem same amount of water. Now in both cases their mass same anyhow latent heat of water is constant same and in both cases that coil is connected to same means voltage same M l by V² is a constant that means I can write time proportional to resistance The time taken to boil water in this problem is proportional to resistance.

[01:14:47]Then ts2 by t1 is r2 by r1 t2 by t1 t1 t1 where is t1 12 minutes r1 is giving percentage now so I'm taking 100 r1 20% of the turns of coil cut and removed means remaining coil length is 80%.

[01:15:24]Length 80% means resistance will be 80%.

[01:15:30]So this will be 4 by 5.

[01:15:35]So t2 is 12 into 48 by 5 9 something you'll get. Yes, this is answer.

[01:16:00]Now some points briefly I want to tell you about this power.

[01:16:08]See one single coil when it is used coil A when it is it's of resistance R1 when A is used time to boil till water is t1 they will give this like this another model I'm telling and another coil B is there its resistance is R2 when B is used Time to boil water is t2. This they'll give individually.

[01:17:14]When a is used, time to boil is t1. When B is T2, when both are used in series, When both are used in series, then the time to boil water is dash.

[01:17:42]Or when both are used in parallel, what is the time taken to boil the same amount of water like this? Another important is there on power.

[01:17:55]This how to do is what is changing here?

[01:17:58]When one resistance is R1. Next R2 when they are connected in series R1 + R2 the effective resistance resistance changing voltage is same. So again you have to use that only ml = v² into t by r here r and t not changing r and t not changing what I'm doing is r formula I'm writing from this v² t by ml this v² by ml whole thing is a constant in this process in any all the cases. So I'm writing some K R is KT I'm writing now when the both the coyles are used in series what is their effective resistance R value R1 + R2 now in place of R KT I'm writing where T is the time taken to boil water when both are used in series KT R1 is KT1 1 R2 is K2 K will cancel T = T1 + T2 you'll get this is the time taken to boil the same amount of water when both the coils are used in series suppose when both the coils are used in parallel when both are used in parallel combination to say mains. What is time taken to boil water is asked. Then what you do in parallel combination effective resistance is 1x r = 1x r1 + 1 by r2 r kt r1 k d1 r 2 k t2 kk cancel 1x t = 1x t1 + 1x t2 t = roo<unk> 2 t1 ts2 by t_sub_1 + tsub_2 this is time taken to boil the water when those two coils are simultaneously used in parallel combination this also very important model power power problems these are different models important I'm just discussing and you remember one more point in any electric circuit The power dissipated will be maximum when resistance in external circuit and internal resistance of cell are equal.

[01:20:59]Power dissipated is maximum.

[01:21:08]Current also current also maximum power is maximum when external resistance in the circuit and internal resistance of cell are equal.

[01:21:20]Suppose they get like this one 6 ohm resistor and 3 ohm resistor are connected in parallel to a cell.

[01:21:38]Power in this circuit is maximum.

[01:21:46]What is internal resistance of the cell? They're asking the internal resistance of this cell must be equal to the effect to external resistance. In the external circuit 6 and three are in parallel. their effective value 63 paral R1 R2 by R1 + R2 if you apply 2 ohm you will get external resistance internal resistance also must be two these are important points okay students I wish you all the best we'll meet in the next session All these notes will be sent to you in PDF form.

[01:23:22]Okay.

[01:23:34]Students, all these notes will be sent to you in PDF form. Okay.