AP Physics 2: 2.4 Changes and Conservation Laws

it's a squishy bag with a countdown timer and it realizes it's a gravy [A bag of TNT appears]

grenade sure she could run away and save herself

but Hannah thinks she knows enough about electric circuits to defuse this thing

less than 10 seconds left anna cuts the wire embraces herself [Hannah cuts wire]

two-one-zero nothing happens the gravy grenade has been defused she jumps up

with a booyah and then a mound of mashed potato it's her right between the eyes [Mashed potato strikes hannah in the face]

take a look at this diagram of the circuit right there yeah all right

what's the circuit I won in a gravy grenade circuit and your essential

answers all right well seems like there's an arms race and food fight [Men throw pie into face]

these days if there are gravy grenades well we're just a few steps away from a

butter bazooka which could be devastatingly slippery [Man with butter bazooka]

luckily we can work on disarmament maybe get some kind of cafeteria peace prize

from the lunch lady well this circuit has two loops which

means we'll be employing the loop rule the loop rule basically states that when

following any closed loop around a circuit there is no change in the

potential there are a few rules that go along with this one when crossing the

resistor in the same direction as the current the potential drops by current

times resistance 2 when crossing the resistor in the opposite direction to

the current the potential increases by current times resistance 3 when moving

through a battery from negative positive the potential increases by voltage 4

when moving through a battery from positive to negative the potential

decreases by voltage well in this circuit we have two loops first we have [loops of circuit highlighted]

AFEBA let's break this loop down step-by-step keeping in mind the rules

we just covered from A to F we move through v1 from negative to positive

which adds positive c1 from F to e we cross our one in the direction of i1

which adds negative I,1 r1 from E to B we cross our 4 in the direction of

I 1 plus I 2 which adds negative I 1 plus I 2 times R 4 from B to a there is

no change the changes in potential along the loop sum to 0 well here's the whole

thing it's an equation walking out way through the loop CDEBC we'll find an

equation that looks like this there we go okay we were given the [Equation appears on circuit]

voltages and resistances so we can plug those numbers into the equations

well we can simplify those equations and set them as simultaneous now we're

getting out to break up in here to solve this we multiply the AFEBA loop by 3

making it easier to deal with the second variable Omega I to remember we're

solving I sub 1 we then subtract loop C de BC from this equation [Equation appears]

leaving us with a new equation of a hundred sixty ohms times current I sub 1

equals 20 volts since current equals volts over resistance we can divide the

voltage 20 by the resistance 160 to find that current one equals one eight

amperes meaning the correct answer is a and hopefully everyone has learned their [Man carrying butter bazooka]

left and when it comes to food fight you know that we're never going to use food

as ammunition again but then again we're not sure what they feed us in our

cafeteria actually counts as food [Man turns green after sniffing cafeteria food]