Parallel Resistance

Resistors in parallel may take a few more calculations.

When there are parallel paths, some current will flow through one path and some through another path(s). The voltage across a group of parallel items is the same for each item. Ohms Law tells us that for any given voltage across a circuit, or section of a circuit, that voltage will equal the product of the circuit current, times the circuit resistance. If the current increases, then the resistance must decrease by some amount to keep the product the same. So in the formula E= I x R the current is inversely proportional to the resistance. We also know that if there are two resistors of different values in parallel, the path with the lower value of resistance to the current flowing will have the more current flowing through it.

With resistors in parallel we need to calculate the circuits equivalent resistance before we can calculate the total current flowing through the set of resistors. There are two formulas for finding the total resistance of parallel resistors in a circuit. A shorthand formula for just 2 resistors in parallel and the general formula for any number of resistors in parallel circuit.

In this circuit diagram there are two resistors labeled R1 and R2 wired in parallel. Some of the current will flow through R1 and some other current will flow through R2. One question that comes to mind is how much current flows through each path? The voltage across each resistor is the same.

 Given:  9 volt supply, R1 is 510 Ohms, R2 is 1000 ohms.
 Find: Current flowing through the R1 path and through the R2 path?

 Step 1: Find the total resistance.
   SHORTHAND FORMULA: (works for just two resistors)
      (R1 X R2) / (R1 + R2)  The product, divided by the sum.
      (510 X 1000) / (510 + 1000 ) OR (510,000/1510) OR 337 ohms

   GENERAL FORMULA:
      1 / ( (1/R1) + (1/R2) + and so on (1/Rx) )
    Applying the general formula to this problem - R1 and R2
    ( 1 / ( 0.00196 + .001 ) )  OR (1/0.00296) OR 338 Ohms

   NOTE: The Total resistance of resistors in parallel will always
         be less than the smallest value resistor.

 Step 2: Find the total current.   I = E / R  
       I = 9(V) / 337(Ohms) =  0.0267 A, rounded to 27 ma

 Step 3: Find the current through each path
         Percent = Total Resistance / resistor value.
           For R1   337 /  510  = 66 %
           For R2   337 / 1000  = 34 %

         Total current times percent of current or
           .027 X .66 =  rounded to 0.018A or    18 ma for R1
           .027 x .34 =  rounded to  .009A or     9 ma for R2

    Take a moment to think about these answers.  
    One path has twice the resistance to current flow, so it
    is reasonable that it will only have half the current flow.

In this circuit diagram three resistors labeled R1, R2 and R3 wired in parallel. In this circuit the current is split into three paths.

 Given:
  9 volt supply
  R1 is 1000 Ohms or 1k ohm
  R2 is 2700 ohms or 2.7k ohm
  R3 is 3300 ohms or 3.3k ohm
 Find: Current flow in each path: for R1, R2 and R3.

 Step 1: Calculate the Total Resistance 
    General formula:
    1 / ((1/R1) + (1/R2) + (1/R3))
    1 / (0.001 + 0.00037 + 0.0003) OR 1/0.00167 OR 598.8 Ohms

 Step 2: Calculate Total Current
    General Current Formula: I = E / R 
          9(volts) / 600(ohms rounded off) = 15 ma

 Step 3: Find the current through each path
         Percent = Total Resistance / resistor value.
         For R1  598 / 1000 = 60 %
         For R2  598 / 2700 = 22 %
         For R3  598 / 3300 = 18 %

         Path current
         R1   0.015 x .60 = 9 ma
         R2   0.015 x .22 = 3.3 ma
         R3   0.015 x .18 = 2.7 ma

  Check the work we just did:
  R1 is the lease resistance and the most current.
  R2 and R3 currents seem reasonable.

Combination Circuits

Resistors in combination circuits take more time but are not all that hard to calculate. Bringing series and parallel circuits all together they function as one combination circuit. The process is to simplify and redrawn, over and over until it is solved.

 
 Given:  9 volt supply, All 6 Resistors are 1000 Ohms
 Find:   Voltages at test points A, B and C with respect to ground.

   NOTE: With the voltage and all resistor values given, the total
         R-equivalent can be calculated.  With those totals the
         total current can be calculated.
   
 Here are the steps.
  Step 1:
    
   Solve the parallel set R5 and R6
   Work: R56 = (1000x1000) / (1000 + 1000) =  500 Ohms
   R-equivalent  R56 = 500 Ohms
   After redrawing the circuit with the new R56
     resistor in place, it should look like this. 



 Step 2: 
    
   Solve the series set R4 and R56
   Work R456 = 1000 + 500 = 1500 Ohms
   R-equivalent  R456 = 1500 Ohms
   After redrawing the circuit with the new R456
     resistor in place, it should look like this. 


 Step 3: 
    
   Solve the series set R2 and R3
   Work R23 = 1000 + 1000 = 2k Ohms
   R-equivalent  R23 = 2k Ohms
   After redrawing the circuit with the new R23
     resistor in place, it should look like this. 


 Step 4:
    
   Solve the parallel set R23 and R456
   Work R2-6 ( 1500 x 2000 ) / (1500 + 2000 ) = 857 ohms
   R-equivalent R2-6 = 857 Ohms
   After redrawing the circuit with the new R2to6
     resistor in place, it should look like this. 


 Step 5: Solve the series set R1 and R2-6 and this 
         is the real R-equivalent across the source power supply.
         Work: Total R = 1000 + 857 = 1857 Ohms


 Step 6: Find the total circuit current.
         Work: Current = 9 V / 1857 R = 4.85 mA


 Step 7: Find the voltage at point A with respect to the battery
         minus terminal (-)
         Work: Supply voltage minus R1 voltage drop.
               9V – ( 1000 X 0.00485) = 4.15 V


 Step 8: Find the voltage at point B with respect to the battery
         minus terminal (-)
         Known: Voltage at point A is 4.15 volts.
                R23 is a voltage divider across the 4.15 volts.
         Voltage at point B is half voltage at point A or 2.075 volts


 Step 9: Find the voltage at point C with respect to the battery
         minus terminal (-)
         Known: Voltage at point A = 4.15
         R4 and R56 is voltage divider across the 4.15 volts.
         R56 is one third the resistance of the divider.
         Voltage at point C =    4.15(V) times .33 = 1.39 V
   Option 2: Using R456 and voltage A,
         calculate current through R456, then voltage across R56

Going To The Work Bench

Experiment - The Voltage Divider:

NOTES:
1) Do not build this project unless you have a volt meter to verity your voltage readings.
2) Do not leave this circuit connected for an extended period of time as it will discharge the battery.
Build this experiment, following the instructions that accompany the kit. Review the Resistor Color Code chart and wire up the kit as described. Perform the lab, and record your results.

                 Parts layout:
                         R4    
                    R1          R5
                         R2     R6
Parts List:                   R3
  1 9-volt battery
  1 Battery power clip
  6 1k ohm resistors - R1, R2, R3, R4,
      R5, R6   (brown, black, red)
  1 experimenters board
  Miscellaneous:
    hookup wire

Short Circuit and Open Circuit:

A short circuit is when a section of a circuit fails due to a component having much lower then expected or zero resistance. A short circuit can be simulated by placing a wire jumper across a component. Many times circuits are not designed to allow shorts so creating one is a poor idea.
In this book I have designed a few safe circuits where we will experiment with planned short circuits We will use this concept to replace a switch as we move along in the book.

Open circuits occur causing a circuit fails to work properly because a component does not behave as if it is in a circuit. The component is called open.

In this section you will start building your trouble shooting skills. You can either just run through the math or you can use the Work Bench trainer and substitute R6 and measure the results as indicated in the problems. Either way we will be using the combination circuit just discussed.

 If R6 became a short circuit, or zero ohms on the diagram.
   On the trainer, remove R6 and replace it with the jumper wire.
   
   Find new voltage at Point A.
   Find new voltage at Point B.
   Find new voltage at Point C.

 If R6 became an open circuit or just remove it from the diagram.
   On the trainer, remove R6 or the jumper wire that replaced
   R6 in the previous question.
   
   Find new voltage at Point A.
   Find new voltage at Point B.
   Find new voltage at Point C.

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