Note: The problem is much easier when one is familiar with what a series vs. parallel circuit looks like. Fortunately, plenty of examples can be found on the internet.
Q: Two resistors are submitted to a 12 V potential. When linked together, the current is of 1.33 A. When in parallel, the current is 5.4 A. What are the values of the resistances?
A) 4 OHM and 5 OHM
B) 4 OHM and 2 OHM
C) 7 OHM and 2 OHM
D) 5 OHM and 1 OHM
E) 4.5 OHM and 4.5 OHM
A: In a series circuit, the total amount of joule energy ( J ) provided by the current ( I ) in quantities of one coulomb ( C ) is represented by Voltage = ( V ).
It can be mathematically proven that V = ( I )( R ), where I = current in units of Amperes, and R=resistance in units of Ohms. Thus, the total Voltage ( V ) of the system = ( I total )( R total ).
Series circuits are sometimes referred to as ” voltage splitters “. The total quantity of electrons passing through a series circuit pass through each resistor ( R ) in the circuit. At each resistor, a voltage drop of IR occurs. Since the current ( I ) is the same through each resistor, the values of R will determine the voltage drop that occurs at each resistor, and the sum of these voltage drops = total voltage.
Thus V total = V1 + V2. Additionally, ( I )( R total ) = ( I )( R1 ) + ( I )( R2 ). Factoring out I on the right-hand side of the equation gives us ( I )( R total ) = ( I )( R1 + R2 ). Dividing each side by I gives us the equation for the total resistance of a series circuit: ( R total ) = ( R1 + R2 ).
Now, getting back to the question 
. I’m assuming that there are two resistors ( 2R ) that equally resist the flow of electrons in the circuit. ( 1R + 1R ) = ( R total ) = 2R.
Therefore, ( V total / I total ) = ( 2R ). Solving for R gives us ( R ) = ( 1 / 2 )( V total / I total ). Plugging the values above into the equation gives us ( R ) = ( 1 / 2 )( 12V / 1.33A ) = ( 6V / 1.33A ) = 4.5 Ohms.
B: In a parallel circuit, the total sum of current ( I ) does not travel across both resistors ( R ). Therefore, a parallel circuit can be thought of as a current divider. Thus, ( I total ) = ( I1 + I2 ).
Since V = IR, I = ( V / R ). Additionally, ( I total ) = ( I1 + I2 ) = ( V total / R total ) = ( V1 / R1 ) + ( V2 / R2 ).
Now, here is what oftentimes causes students trouble when dealing with parallel circuits. The amount of current ( I ) crossing each resistor in parallel may or may not be different, but the voltage drops ( V ) at each resistor in parallel is the SAME. For example, if there are two resistors that happen to have different resistances in parallel, the one with the least resistance ( R ) will have more current passing over it in a unit of time, but the energy drop PER COULOMB of charge is the same. If it takes $100.00 to purchase 100 candy bars from a store, one store may sell out more quickly than another, but for every $100.00 spent, 100 candy bars are sold in both stores.
( I total ) = ( I1 + I2 ). Therefore, ( V total / R total ) = ( ( V1 / R1 ) + ( V2 / R2 ) ). Since the voltage drops are the same, V1 = V2, and V can be factored out of the fractions on the right hand side of the equation. Therefore, ( V total / R total ) = ( V )( ( 1 / R1 ) + ( 1 / R2 ) ). Dividing both sides by V gives us the familiar expression for voltage ( V ) in a parallel circuit: ( 1 / R total ) = ( ( 1 / R1 ) + ( 1 / R2 ) ). Solving for R will require us to take the reciprocal of the value we get from adding ( ( 1 / R1 ) + ( 1 / R2 ) ).
( V total / I total ) = ( R total ) = 2.2 Ohms. ( 1 / 2.2 Ohms ) = ( 0.45 / Ohms ).
( 0.45 / Ohms ) = ( ( 1 / R1 ) + ( 1 / R2 ) ).
Using cross multiplication, we obtain a common denominator for our fraction, and ( 0.45 / Ohm ) = ( ( R1 + R2 ) / R1*R2 ). If R1 and R2 are equal to one another ( and we must make that assumption if the R values are not given ), then ( 0.45 / Ohm ) = ( 2R1 / ( R1*R1 ). Thus, ( 2R1 / R1^2 ) = ( 0.45 / Ohm ). ( 2 / R1 ) = ( 0.45 / Ohm ). Solving for R1, we get R1 = ( 2 / ( 0.45 / Ohm ) ). R1 = R2 = 4.44 Ohms.
RENAISSANCE PHYSICS 