hello everybody my name is Iman welcome back to my YouTube channel today we're going to do problems associated with our circuit chapter let's go ahead and get started one says if a d for defibrillator passes 15 amperes of current through a patient's body for 0.1 seconds how much charge goes through the patient's skin now electrical current is defined as charge flow or in mathematical terms charge transferred per time we could look take a look back at our equation for that I equals Q over delta T So current equals charge over change in time now a 15 ampere current that acts for 0.1 seconds how much charge is that going to transfer well let's rearrange this equation to solve for charge considering we know current and we know time so if we rearrange this equation we can say that charge is equal to current times time and we have the amount of current 15 amperes we have the time 0.1 seconds and this gives us a charge of 1.5 coulombs all right so a current that acts for 0.1 second will transfer 1.5 coulombs of charge that is going to be answer Choice B fantastic two says a change of two micro coulomb flows from the positive terminal of a six volt battery through a hundred Ohm resistor and back through the battery to the positive terminal what is the total potential difference experienced by the charge now kirchhoff's Loop rule it states that the total potential difference around any closed loop of a circuit is zero volts all right this is one of the Cur this this is one of the kirchhoff's rules that we covered we covered his Loop Rule and his Junction Rule and his Loop rolls Loop roll Loop rule sorry guys states that the total potential difference around any closed loop of a circus circuit is zero volts now another way of saying this is that the voltage gained in the battery 6 volts is going to be used up through the resistors and because this charge both started and ended at the positive terminal its potential difference is just going to be zero volts all right so the answer here is just based off of kirchhoff's loop rule zero volts you don't have to do any math here you just have to remember your circuit laws and rules all right fantastic three says the resistance of two conductors of equal cross-sectional area and equal length are compared and there are found to be in a one to two ratio the resistivities of the materials from which they are constructed must therefore be in what ratio all right so the resistance of a resistor is given by this formula resistance is equal to resistivity times length time over cross-sectional area there is a direct proportionality between resistance and resistivity all right because the other variables are equal between the two resistors we can determine that if R1 R2 is in a one to two ratio all right if if we know that the resistive the resistance of the two conductors all right of equal cross-sectional area equal length if they have a one to two ratio for R1 R2 and we know that resistance and resistivity here this row R linearly proportional then the resistivities have to be in the same ratio as well all right and so the correct answer here for 3 is B one to two ratio Fantastic Four says a voltaic cell provides a current of 0.5 amperes when in a circuit with a 3 Ohm resistor if the internal resistance of the cell is 0.1 ohms what is the voltage across the terminal of the battery when there is no current flowing all right so the question is going to be testing our understanding of batteries in a circuit the voltage across the terminal of the battery when there is no current flowing this is referred to as our electromotive Force all right our electromotive Force EMF however when a current is flowing through the circuit the voltage across the terminals of the battery is decreased by an amount that's equal to the current multiplied by the internal resistance of the battery we covered this concept and we displayed it mathematically with this expression right here now to determine the EMF of the battery we're first going to have to start by calculating the voltage across the battery when the current is flowing and the best way to do that first and foremost is using Ohm's law so we can figure out that voltage we have the current we have the resistance all right we're just going to plug those in 0.5 amperes multiplied by 3 ohms this gives us a voltage of 1.5 volts now because we know the internal resistance of the battery and the current and the voltage now we can use this expression all right we can use this expression now we're going to shift things around so that we can solve for the electromotive force the EMF right here all right the E all right this is going to be equal to voltage plus i r right we know what the voltage is it's 1.5 volts we just calculated that calculated that all right plus the internal resistance we were given that 0.5 amperes our um and one and 0.1 ohms we were given both these values the current and the internal resistance okay now we multiply this out what we're going to get is 1.55 volts and this answer it makes sense in the context of a real battery because its internal resistance is supposed to be really small and so that the voltage provided to the circuit is pretty much as close as possible to the EMF of the cell when there is no current running and so the correct answer here is going to be D 4 is d fantastic five says a Transformer is a device that takes an input voltage and produces an output voltage that can be either larger or smaller than the input voltage depending on the Transformer design of course although the voltage is it although the voltage is changed by the Transformer energy is not so the input power equals the output power a particular Transformer it produces an output voltage that is 300 percent of the input voltage what is the ratio of the output current to the input current okay so that's that's a lot of information and we're told here that Transformers can serve energy so that output power equals input power now because power input power equals output power all right we can change this power term to um to its definition in terms of the variables current and voltage so if you remember we discussed that power had a couple of of Expressions that we can write based off of what variables you're looking to express it in so one one expression for power is that power equals current times voltage okay and that's the one that we're going to use but just to remind you remember we said power equal to current times voltage or it can be equal to current squared times resistance or it could be equal to voltage squared over resistance and you can get any of these relationships from Ohm's law all right now what we're told here is power n equals power out and then we're given this information that um that the particular Transformer produces an output voltage that is 300 of the input voltage and then we want the ratio of the current so we obviously want to use an expression of power that equals current and voltage because those are the two variables we're using and and trying to figure out for this problem okay so if power n equals power out then current times voltage n should equal current times volt voltage out all right so we can work from here all right so there is therefore an inverse proportionality between current and voltage so if the output voltage let's highlight that in red if the output voltage is 300 of the input voltage so what that means is it's three times the input voltage all right then the output current must be one-third of the input current all right so what this this represents is a one to three ratio let's visualize this if the output is 300 percent more than the voltage input then what we can say is i n v i n is equal to I out three times V in all right now if this is equal to 3 v i n current out and that's equal to current and voltage in all right if this is if this term has a 3 right here associated with it all right then this output current to balance it out so that this statement is still true that in equals out then this current term has to have a one-third associated with it all right so that this can be equal all right and so if the output voltage is 300 of the input voltage then the output current has to be one-third of the input voltage all right so that things can cancel out and so what that means is current out has to be one third of current in and so that way when we write this out current n equals a current n times voltage is going to be equal to 3 voltage in multiplied by 1 3 current in all right all right this is equal to this and this way this statement is indeed true because these three and one-third cancel each other out all right this is a one to three ratio so the correct answer for 5 here is going to be a beautiful 6 says given that R1 is equal to 20 ohms R2 is 4 R3 equals R4 and they both equal 32 ohms R5 equals 15 and R6 equals 5 ohms what is the total resistance in the setup shown below okay the fastest way to tackle this kind of question is to simplify the circuit bit by bit so for example notice that R3 and R4 are in parallel with each other and then they are in series with R2 all right and then similarly R5 and R6 are in series all right they are in series right here all right and if we determine the total resistance in each bandage we will just be left with three branches in parallel all right so if we tackle those like that way it's going to be a lot easier okay so to start we're going to find the resistance in the middle Branch this guy right here we're going to figure this guy first out and then what we really have is just these three branches of that that look like a parallels all right so we're going to tackle this middle Branch first and foremost R2 R3 R4 all right starting off with R3 and R4 this middle Branch these are in parallel all right and so if we're trying to find the total resistance of R3 plus R4 it's just going to be 1 over R3 plus 1 over R4 because they're in parallel so remember the relationship for resistance for parallel it's 1 over and it's in additive that way all right so remember to recall your total table I'm going to go back to the notes here to remind us right please watch the lectures because we go in a lot more depth but here is the table that we had that summarizes our relationship between current resistance voltage and capacitance for series and parallel circuits so for resistance for parallel all right we have our relationship right here so this is what we're using to first and foremost figure out R3 and R4 which are in parallel to each other all right this is going to give us 1 over 32 plus 1 over 32 which is equal to 2 over 32. all right 2 over 32 and then what we want to do is we're going to do the inverse of each side so that we can find the Resistance 3 plus 4 and that's going to be equal to 32 over 2 which is equal to 16 ohms so we figured that out now we want to look at the bigger picture now that we've figured out three uh R3 plus R4 all right and we think of it as one kind of resistor here it's kind of in series with R2 so we're going to figure it this whole thing out and out together uh in series so our two plus three plus four is just going to be R2 plus r three plus four here all right and that's going to be 4 plus 16 which is 20 ohms perfect all right now the circuit Can Be Few viewed as three resistors in parallel after we figure out what R R5 and R6 are so that's another thing we want to figure out R5 plus 6 is just um 15 plus 5. all right fifteen plus five which is 20 um now we can think of this as a series all right a parallel uh now we can think of it as a circuit a parallel circuit all right where this is now R5 plus R6 this is our two plus three plus four and this is R1 and we have all this information R 5 plus 6 is 20 ohms all right r two plus three plus four is also 20 ohms and then R1 is simply 20. all right so now we can find the total resistance by treating this as a parallel circuit all right and we know what this looks like for resistance calculations for a parallel circuit it's going to be R total okay is going to be 1 over R1 plus 1 over r two plus three plus four plus one over r five plus six and this is just going to be 1 over 20 plus 1 over 20 plus 1 over 20 and that gives us 1 over R total uh equal to 3 over 20. all right then we're going to do the reciprocal of both side so our total is equal to twenty over three which is about 6.7 ohms all right that's going to be closest to answer Choice B so 6 here is B fantastic that's a lot of work but we did it we got through all right now let's read number seven seven says how many moles of electrons pass through a circuit containing a hundred volt battery and a 2 ohm resistor over a period of 10 seconds and look at that we're given what are ferret a farad unit is beautiful so to determine the mole of charge that passed through the circuit over a period of 10 seconds we're going to have to calculate the amount of charge that's actually running through the circuit all right and charge is simply current times time all right that means we're going to have to figure out what you know and then the current can be calculated all right it can be calculated using Ohm's law all right so let's write all our expressions for this right here okay so we know that voltage is equal to uh current times resistance and we know that current is equal to charge over change in time all right now we can shift this guy around to solve for charge charge is equal to current over current times time all right current times time but then go ahead and replace current with a different term all right let's use our Ohm's law all right and figure out a different definition for current current is voltage over resistance so let's replace current with that term right here and that means now that charge is equal to voltage times time over resistance we have these values so we can start to plug them in all right voltage 100 volts 10 seconds resistance of 2 ohms this gives us 500 coulombs that we're not done yet right we need to calculate the number of moles of charge okay and that's going to be represented by using the Faraday constant and approximating f as 10 to the 5 coulombs per mole um a mole of electrons all right so we're going to take this 500 coulombs we're going to convert it using some dimensional analysis all right 10 to the 5 coulombs is approximately one mole of electrons fundamental charge all right electrons all right and that's going to give us 5 times 10 to the minus 3 moles of electrons all right this is going to be the closest to answer Choice a here so 7 is a fantastic eight says in the circuit below what is the voltage drop across the two over three Ohm resistor so this guy right here now to determine the voltage drop across this resistor we're going to have to start by calculating the total resistance in the circuit for the resistors in parallel all right for the resistors in parallel um so these guys right here all right the equivalent resistance is going to be determined using this expression right here right 1 over the total resistance of these two is going to be equal to 1 over the two ohms all right plus one over two over three all right and that's going to give us uh two right let me see actually I can simplify this a little better so that we can see this one over two over three is just 3 over 2. 2 all right that gives us 4 over 2 which gives us 2. all right so that gives us two but let's do the inverse of both sides now so that we can figure this out all right and that gives us one half ohms all right so that's the resistance for that now the total resistance in the circuit is the sum of the remaining resistors all right so if we figured out what this resistor is as a in parallel all right now we can think of this and this as two resistors in series all right and then we can find the total resistance in the circuit using our our resistance formula for series circuits and that's just going to be the sum of all the resistors and so now we can find the resistors uh by adding them so that's going to be one half ohms plus one half and that gives us one one ohm all right now that we know the equivalent resistance we can finally calculate the total current using Ohm's law right current is equal to voltage over resistance right from from Ohm's law which is V is equal to current over resistance I just shifted things around here all right because we're trying to solve for current all right we can just plug in these values 10 volts over one ohm and that gives us 10 amperes all right so the answer is 10 amperes all right oh it seems like my answer choices have been cut off but the answer is 10 amperes beautiful that's eight all right let's keep going nine if the area of a capacitor plate is doubled while the distance between them is half how will the final capacitance compare to the original capacitance so this question should bring to mind our equation that capacitance is equal to the permittivity of free space multiplied by area over distance all right a is area of the of the place D is the distance between those plates from this information and this equation we can infer that if we were to double the area we're going to also double the capacitance all right um if we double the area we're going to double the capacitance there's a direct relationship between capacitance and area they're both in the numerator right all right so if you double the area you're going to double the capacitance okay let's look at the same equation now and see what happens if we play with the distance all right if the distance is halfed all right what if we go ahead and cut this by half or multiply this by half what will happen well the distance and the capacitance are inversely proportional notice that the capacitance is in the numerator right if you think of everything as over one all right and the distance this is in the denominator all right so it has this inverse relationship if we have the distance all right then we multiply the capacitance by the reciprocal of that so we multiply we double the capacitance if we were to half the distance so what if we're doing both at the same time if we double the area and we have the distance all right then we are changing the capacitance by a factor of four the new capacitance is four times larger than the initial capacitance so the correct answer here is answer Choice D all right fantastic 10 says the energy stored in a fully charged capacitor is given by this equation U equals one half CV squared in a typical cardiac defibrillator a capacitor charge to 7500 volts has a stored energy of 400 joules based on this information what is the charge on the capacitor in the cardiac defibrillator okay very good question now the question is asking us to calculate the charge on the capacitor so we can use our our expression Q equals CV all right we've seen this before as you know C is equal to um Q over V right but we have we have C we have capacitance and we have voltage so we've shifted around things so that we can solve for charge here that's where we got this equation we're given that V is 7500 volts we can calculate and we can calculate capacitance from the formula for energy which is U equals one half CV squared all right so what that excuse me what that means is that in this equation you equals one-half CV squared we're going to replace C with Q over V here all right so then U is equal to one half Q over v v squared this V and one of these V's cancel out so now we have one half q v all right now Q is equal to if we shift things around 2 U over V all right now we have all this information so we can solve it that's why we're doing all this manipulation because we only have some information here all right we have we have um U and we have voltage so we took our initial understanding of capacitance equals charge over voltage we took the formula they gave us in the problem U equals one half CV squared we replaced capacitance because we don't have that information in the problem with charge over volts things canceled out and then we just shifted things around now that we know U is equal to one half QV we simply shifted things around to solve for Q so we multiplied both sides by 2v 2 over V all right and 2 over B and so now we could solve for charge because we know what U is and we know what V is and we can plug them in now so now 2 times 400 joules over 7500 volts that gives us about 0.1 coulomb so the charge is close to 0.1 coulomb and that's and the answer to this that's closest to that is going to be answer choice so 10 is C beautiful 11 says excuse me guys a 10 Ohm resistor carries a current that varies as a function of time as shown how much energy has been dissipated by the resistor after five seconds beautiful now power power is energy dissipated per unit time therefore the energy dissipated okay so power is energy dissipated over time we can shift this around to Define um to Define energy if power is equal to energy dissipated over time then energy is equal to Power Times time all right times change in time in the five second interval during which the resistor is active it has a 2 ampere current for three of those seconds the power dissipated by a resistor carrying the current I is figured out by this equation Powers equal to current squared times resistor therefore the energy dissipated we can figure out by combining these two expressions right here we can replace p with current squared over R and so change in energy is equal to I squared r times delta T we have this information current is two amperes squared resistors 10 ohms and change in time is the three seconds all right this is going to give us 120 joules so the correct answer here is C beautiful 12 says in the figure below six currents meet at Point p all right what is the magnitude and direction of the current between point p and X all right so this is six amp amperes away five amperes away um and then we have two amperes two point P eight amperes towards p and three amperes towards p what we're gonna have to use here is karkov's Junction rule which states that the sum of all currents directed into a point is always equal to the sum of all currents directed out of the point now the currents directed into the point P are a two and three so this this and this so eight amperes plus three amperes plus two amperes that's all of the current in that has to equal all of the current out all right now the currents the current directed out or just what we see here five and six so they're going to be equal to five amperes plus six amperes all right that gives us this is about 13 on this side and that is about 11 on this side all right now because the two numbers have to equal each other all right there has to be something that's added on this side to make it equal to 13. we can add plus two amperes here and then the total would be 13 on both sides all right now we added this 2 ampere on the side of current out so that is what x is X is 2 amperes and it's pointed away from P towards X all right that's going to be answer Choice a fantastic 13 says which of the following will most likely increase the electric field between the plate of a parallel plate capacitor let's read the answer choices adding a resistor that's connected to the capacitor in series B says adding a resistor that's connected to the capacitor in parallel C says increasing the distance between the plates and D says adding an extra battery to the system now the electric field between two plates of a parallel plate capacitor is related to the potential difference between the plate of the capacitor and the distance between the plates this is Express this is this can be expressed mathematically we've seen this Formula E equals V over D now the addition of another battery it's going to increase the voltage applied to the Circuit all right so it's going to increase that voltage which would also increase the electric field all right the addition of a resistance series that will increase the resistance and then decrease the voltage applied to the capacitor all right and so that means answer Choice a is not a very good answer here all right adding a resistor in parallel it will not change the voltage drop across the capacitor and it should not change the electric field either so that cancels out answer Choice B what about increasing the distance increasing the distance between the plates would decrease the electric field not increases so that's also incorrect all right the best answer Choice here is adding an extra battery to the system all right adding an extra battery will increase the voltage applied to the circuit and that's going to increase the electric field beautiful 14 says each of the resistors shown carries an individual resistance of 4 Ohms assuming negligible resistance in The Wire what is the overall resistance of the circuit now the resistance of the three resistors wired in a series these right here is just going to be the sum of the individual ones each of them is four so four plus four plus four that's equal to 12. and now if we think of this as one resistor that has 12 ohms it's in parallel with this one right here and then we can determine the overall resistance using our formula for resistance sum of that in a parallel circuit so 1 over R is going to be 1 over R1 plus 1 over r two that's going to be equal to 1 over 12 plus 1 over 4. all right that's going to give us 4 over 12. all right now let's do the inverse of that all right if we do the inverse of that we get RP is equal to 12 over 4 or 3 ohms so the correct answer here for 14 is d last but not least 15 which of the following best characterizes ideal volt meters and ammeters all right now let's read the answer choices a says ideal voltmeters and ammeters have infinite resistance all right this is not true B says ideal voltmeters and ammeters have no resistance also not true C says ideal voltmeters have infinite resistance and ideal ammeters have no resistance this is a correct statement and this is ideal voltmeters have no resistance in ideal animators have infinite resistance incorrect it's the opposite all right so voltmeters are attempting to determine a change in Potential from one point to another and you know to do this they should not provide an alternate route for charge flow and they should therefore have infinite resistance ammeters on the other hand they attempted to determine the flow of charge at a single point and they should not contribute to the resistance of a series circuit so they have they should have no resistance at all which is why the correct answer for 15 is and with that we end our practice problem set let me know if you have any questions comments concerns down below other than that good luck happy studying and have a beautiful beautiful day future doctors