INTRODUCTION TO DIRECT CURRENT ( DC ) ELECTRONICS :
Note: Assuming each resistor ( R ) = 64 Ohms, what is the equivalent resistance of the circuit from points A to B ?
Symmetry regarding the flow of electrons is the key to calculating the total resistance ( R ) of this circuit. Let’s suppose the electrons flow from A to B.
Moving from left to right, the initial currents ( I one ) coming out of A will first go through a set of identical resistors ( R ) positioned at the top and bottom of the circuit. These two identical currents then both arrive at nodes that are physically identical. The currents each make a split equal in magnitude at each node, because the physical situation downstream to both are identical. Let’s call the currents leaving these nodes ( I two ).
Downstream of the first node pair are four energetically identical paths for current to flow. All four currents ( I two ) travel across resistors that are situated parallel to one another. After passing the resistors ( where voltage drops occur ), the four ( I two ) currents recombine downstream at two nodes. These nodes are identical to the nodes that preceded the four-way split. Therefore, the currents emanating from these nodes are equal in value to the upstream ( I one ) current pair that entered the first nodes.
WHEN DEALING WITH PARALLEL CIRCUITS, CURRENT DESIGNATIONS SIMPLIFY THE PROCESS OF DETERMINING WHICH RESISTORS ARE PARALLEL TO ONE ANOTHER. ONCE THIS DESIGNATION HAS BEEN MADE, THE DOUBLE RECIPROCAL METHOD OF PARALLEL CIRCUIT ADDITION CAN BE EMPLOYED FOR THE RESISTORS IN QUESTION.
If these resistors were in series with one another, the same electrons would travel through all of them. This clearly is not the case. The first two parallel resistors must be added using the double reciprocal method. The next four parallel resistors must then be added using the double reciprocal method. Finally, the last two parallel resistors must be added using the double reciprocal method.
Carrying out the aforementioned additions gives us three ” equivalent ” resistor values. Equivalent resistor values can be drawn on a straight line and treated as series resistors that are added directly.
Let’s begin by adding the four parallel resistors using the double reciprocal method. First ( ( 1 / R ) + ( 1 / R ) + ( 1 / R ) + ( 1 / R ) ) = ( 4 / R ). Now, we must take the reciprocal of this fraction ( DOUBLE RECIPROCAL ) to get the correct answer. The reciprocal of ( 4 / R ) = ( R / 4 ) Substitute the value of R to get the correct answer. ( 64 OHMs / 4 ) = 16 OHMs for the equivalent resistance of the four parallel resistors.
The pairs of two resistors can be added as ( ( 1 / R ) + ( 1 / R ) ) = ( 2 / R ). Next, take the reciprocal of this answer to get ( R / 2 ). Substitution of R gives us ( 64 OHMs / 2 ) = 32 OHMs each.
We now have three resistor equivalents that are in series with one another. Adding them yields a total resistance of 32 OHMs + 16 OHMs + 32 OHMs = 80 OHMs.
SUMMARY : The sum of the four parallel resistors is ( 1 / R ) + ( 1 / R ) + ( 1 / R ) + ( 1 / R ) = ( 4 / R ). Taking the double reciprocal gives us ( R / 4 ). The other two resistors to the left and right of this equivalent resistance equal ( 1 / R ) + ( 1 / R ) = ( 2 / R ). Taking the double reciprocal gives us ( R / 2 ). We now have three values that can be treated as individual resistors located in series in relation to one another.
Using this method, we must find the lowest common denominator ( LCD ) to add the fractions. Since the LCD = 4, the top and bottom of the ( R / 2 ) rations must be multiplied by 2. Thus, ( R / 2 ) + ( R / 4 ) + ( R / 2 ) = ( 2R / 4 ) + ( R / 4 ) + ( 2R / 4 ) = ( 5R / 4 ). Substituting the given value of 64 ohms into the equation, we have ( ( 5 )( 64 ) / ( 4 ) ) = 80 OHMs.
RENAISSANCE PHYSICS 