We have previously seen how all of the current ( I ) within a series circuit will pass through each resistor ( R ) situated within it. The sum of the energy drops that a coulomb ( C ) of charge loses as it traverses a circuit is equal to the voltage ( V ) value of the source. For this reason, series circuits are sometimes referred to as “ voltage dividers. “ We will now begin our study of parallel circuits. A parallel circuit is one in which conventional current spreads out onto one or more circuit branches after passing a node. Each current then passes through whatever circuit element is situated along them. Finally, the currents all reconvene before traveling to the low-energy-potential side of the source. For this reason, parallel circuits are sometimes referred to as “ current dividers “ :
At this point, we may use ohm’s law to quantify how currents split apart within the circuit:
( V / R ) = I
I1 = I2 + I3
Vs / Rt = V1 / R1 + V2 / R2
The current that crosses each resistor will undergo a voltage drop. How then must we go about determining the voltage drops that occur? It is important to note that current is a measure of the rate at which coulombs of positive charge move past a given point of reference. In order to truly appreciate the relationship ( if any ) between current and voltage within parallel circuits, a thought experiment is due.
Let’s imagine that two coulombs of electrons pass through Node A. One coulomb of charge moves downward across R1, and the other coulomb of charge moves around the circuit until it crosses R2. As each coulomb of charge recombines at node B, would it make any physical sense for one group of electrons to be at a higher energy state than the other? The answer is “ no “. If R1 > R2, there will be more coulombs of charge moving across R2 each second, but each coulomb of charge that crosses Node A and Node B will experience voltage drops that are equal to the source voltage. If either group of electrons approached Node B with a higher energy state than the other, the Law of Conservation of Energy would be violated:
Vs = V1 = V2
We are now ready to derive an expression for the net resistance within a parallel circuit:
Vs / Rt = V1 / R1 + V2 / R2
( Vs )( 1 / Rt ) = ( Vs )( 1 / R1 + 1 / R2 )
1 / Rt = 1 / R1 + 1 / R2
Thus, a “ double reciprocal “ procedure will be used over and over again to determine what resistance exists within one or more circuits that are parallel to one another:
1 / Rt = [ ( R1 + R2 ) / R1R2 ]
Rt = [ ( R1R2 ) / R1 + R2 ]
This formula works when only two resistors are in parallel with one another, but needless to say, many circuits that we encounter will be solved using old fashioned, brute-force mathematics.
RENAISSANCE PHYSICS 