FINDING THE LOWEST COMMON DENOMINATOR:
Let’s first envision putting two half pieces of a pie together to get a full pie. Numerically, this would involve adding ( 1/2 ) + ( 1/2 ) = ( [ 1 + 1 ] / 2 ) = ( 2/2 ) = 1 whole pie. This example was made easy by the fact that we didn’t have to change the bottom portion of the fraction ( denominator ), and we only needed to add the top component of the fraction ( numerator ) to get our answer. In other cases when we’re adding fractions, if the denominators are not the same, we must use algebraic rules to establish a ” common denominator ” before addition of fractions will be possible.
Here’s a great example. Envision adding ( 1/2 ) of a pie to ( 1/4 ) of a pie. It’s easy to envision that we’d end up with ( 3/4 ) of a pie. However, the mathematically appropriate process of obtaining the correct answer would necessitate the use of a common denominator. Let’s attempt to do this problem the incorrect way to see why this is the case.
Incorrect Method: ( 1/2 ) + ( 1/4 ) = ( [ 1 + 1 ] / [ 2 + 4 ] ) = ( 2/6 ) = ( 1/3 ) of a pie. We clearly wouldn’t end up with ( 1/3 ) of a pie if we added ( 1/2 ) of a pie to ( 1/4 ) of a pie.
When adding fractions that have different denominator values [ Ex. ( 1/2 ) + ( 1/4 ) = ? ], finding the lowest common denominator ( LCD ) involves finding the lowest number that each denominator value will divide into evenly. Let’s now add ( 1/4 ) of a pie to ( 1/2 ) of a pie using the correct mathematical technique involving the usage of a LCD.
When evaluating ( 1/2 ) + ( 1/4 ), we first determine what number both denominator values ( ” 2 ” and ” 4 ” ) divide into evenly. We first look at the larger of the two numbers ( ” 4 ” ) and see if the smaller number will divide into it evenly. In this case, we are lucky. The number ” 2 ” divides evenly into ” 4 “, and the number ” 4 ” divides into itself evenly as well. For this reason, the number ” 4 ” is the lowest common denominator ( LCD ). Since ” 4 ” is the LCD, both fractions must have the number ” 4 ” in the denominator in order that they may be added together to obtain a correct answer.
We must now take another algebraic rule into account. ANY NUMBER MULTIPLIED BY ONE RETAINS ITS ORIGINAL VALUE. ( 2 )( 1 ) = 2. Additionally, ( XYZ )( 1 ) = XYZ. Furthermore, since any number divided by itself equals 1, the expression of ( XYZ )( [ 7/7 ] ) = XYZ, because ( 7/7 ) = 1. We will now look again at the problem of adding ( 1/2 ) + ( 1/4 ) and using the above truth to change ( 1/2 ) into a value that has the number ” 4 ” in the denominator.
If the LCD of the fractions ( 1/2 ) and ( 1/4 ) is ” 4 “, we will need to multiply the denominator of ( 1/2 ) by the number ” 2 ” in order to get a denominator with a value of ” 4 “. HOWEVER, WE MUST ALSO MULTIPLY THE TOP OF THE NUMBER ( 1/2 ) BY ” 2 ” IN ORDER TO GET THE FRACTION WE NEED. We’re not really changing the value of ( 1/2 ), but rather, we’re changing it into a ratio that is appropriate for adding it to another fraction. BY MULTIPLYING ( 1/2 ) BY ( 2/2 ), WE END UP WITH A FRACTION THAT HAS THE LCD WE ARE LOOKING FOR.
( 2/2 )( 1/2 ) = ( 2/4 ).
Since we now have the LCD of ” 4 ” in place, ( 2/4 ) + ( 1/4 ) = ( [ 2 + 1 ] / 4 ) = ( 3/4 ). This answer makes sense mathematically and intuitively. ( 2/4 ) of a pie + ( 1/4 ) of a pie = ( 3/4 ) of a pie, and this is essentially the same as adding ( 1/2 ) of a pie to ( 1/4 ) of a pie. The LCD, as we’ve seen, makes the mathematical process of finding a correct answer work in the way that it should.
Let’s now try a slightly harder example. ( 1/2 ) + ( 1/3 ) + ( 1/7 ) = ?. In finding the lowest common denominator ( LCD ), we must find some number that all three denominator values ( ” 2 “, ” 3 “, and ” 7 ” ) divide into evenly. Clearly, neither ” 2 ” or ” 3 ” divide evenly into the number 7. Furthermore, multiplying ” 7 ” and ” 3 ” gives us ” 21 “. The number ” 2 ” cannot divide equally into ” 21 “, so ” 21 ” isn’t the LCD we need. Additionally, multiplying ” 7 ” by ” 2 ” gives us the number ” 14 “. Since the number ” 3 ” cannot divide equally into ” 14 “, the number ” 14 ” is not the LCD we are looking for. At times, finding an LCD can be tedious. For this particular example, however, we can use a little trick to find a number that all three denominators divide into evenly.
If we multiply the numbers ” 2 ” and ” 3 “, we get the number ” 6 “. This solves the problem of finding a number that ” 2 ” and ” 3 ” divide evenly into, but we still need to find some number that ” 2 “, ” 3 ” AND ” 7 ” will divide into evenly. If we take the number ” 6 ” and multiply it by ” 7 “, we get the number ” 42 “. Recall that ” 6 ” is the product of multiplying ” 2 ” and ” 3 ” together, so the first two denominators will divide into ” 42 “, as will the number ” 7 “. We now have a number ( ” 42 ” ) that the denominator values of ” 2 “, ” 3 ” and ” 7 ” will all divide into evenly. Let’s now convert each fraction to an expression that will allow it to be added to the others:
Since ( 7 )( 6 ) = 42, then ( 1/7 )( 6/6 ) = ( 6/42 ).
Since ( 3 )( 14 ) = 42, then ( 1/3 )( 14/14 ) = ( 14/42 ).
Since ( 2 )( 21 ) = 42, then ( 1/2 )( 21/21 ) = ( 21/42 ).
We are now ready to add our fractions together using a LCD.
( 6/42 ) + ( 14/42 ) + ( 21/42 ) = ( 41/42 ).
PARALLEL ELECTRICAL CIRCUITS: For now, let’s skip a conceptual analysis of parallel circuits ( plenty of time for that later ). We will now simply go through the motions of using a LCD to arrive at a correct answer for the total resistance ( R total ) or ( Rt ) of a parallel electrical circuit.
If V=( I )( R ), then the current ” I ” = ( V/R ). The currents, denoted by the letter ” I ” split up within parallel electrical circuits. Therefore, a parallel circuit is often referred to as a ” current divider “. The currents flowing through each path will differ if the resistors ( R ) within each path differ. The total current ( I total ) or ( It ) within the entire circuit can be summed up as follows: ( It ) = ( I1 ) + ( I2 ) + ( I3 ) if there are three separate parallel paths in which electricity is split up. Since I = ( V/R ), we can rewrite the above expression as ( Vt/Rt ) = ( V1/R1 ) + ( V2/R2 ) + ( V3/R3 ).
What we will learn in later studies is that the voltage drop across electrical resistors in parallel is the same ( the currents differ, but regardless of how fast electrons are moving in each pathway [ fast or slow ], each time a coulomb of charges passes the resistor, the same voltage drop has occurred. ) The high current paths will, however, have a higher Wattage ( W ), but that is of no concern at the moment. Since Vt = V1 = V2 = V3, the above equation reduces to ( 1/Rt ) = ( 1/R1 ) + ( 1/R2 ) + ( 1/R3 ).
Example: A parallel circuit has three electrical paths. The first path has a resistance of 2 ohms ( R1 ), a second path has a resistor that is 3 ohms ( R2 ), and the last pathway has a 7 ohm resistor ( R3 ). What is the total resistance of this parallel circuit?
Answer: ( 1/Rt ) = ( 41/42ohms ).
Recall that ( 1/2 ) + ( 1/3 ) + ( 1/7 ) = ( 41/42 ). Since we need the value of Rt, we need to invert the answer we arrived at. Therefore, Rt = ( 42 ohms / 41 ).
Due to having to take the reciprocal twice in order to get Rt of a parallel circuit, the expression for total resistance in a parallel electrical circuit is often written as follows:
Rt = ( 1 / [ ( 1/R1 ) + ( 1/R2 ) + ( 1/R3 ) ] ).
The LCD of the bottom fractions is first determined. Afterwards, the bottom values of ( 1/R ) are added together. Finally, the reciprocal is taken to obtain a value for Rt.
A SIMPLER METHOD:
And now, the easy way 
Let’s assume, once again, that a parallel circuit contains three electrical conductive pathways and that each pathway contains a single resistor. If R1 = 2, R2 = 3, and R3 = 7, we can solve ( 1/Rt ) = ( 1/R1 ) + ( 1/R2 ) + ( 1/R3 ) by multiplying both sides of the equation by the LCD. This will eliminate the fractions on the right hand side of the equation. The LCD = 42, so ( 42 )( 1/Rt ) = ( 42 )( [ 1/R1 ] + [ 1/R2 ] + [ 1/R3 ] ) = the following:
( 42/Rt ) = ( 21 + 14 + 6 ).
We now have ( 42/Rt ) = ( 41 ). The S.I. unit of resistance ( Rt ) is ohms, so Rt = ( 42 ohms/ 41 ). This latter method of adding fractions is a nice method to use after a good grasp of the more rigorous approach has been established.
RENAISSANCE PHYSICS 