Savvy Circuit Parallel Circuit Resistance

Circuit Savvy

Part two:

Parallel Circuit Resistance

In this last article in the Circuit Savvy series we explain the concept of resis-

tance in a parallel circuit and detail

the calculations involved in determining

the total resistance in such a circuit.

Voltage and current in series and parallel circuits are relatively easy to understand because we can use descriptions that are clear and logical. For example, in a series circuit, all components in series pass the same total current (IT) and the sum of the voltage drops around a series circuit equal the total applied voltage (ET). These are easy concepts to grasp using simple addition.

In parallel circuits we learned that all branches have the same voltage across their terminals and total current (IT) divides among the parallel branches. Simple addition or division again.

Editor's Note: Simple mathematics can lead to simple mistakes. The last "Circuit Savvy" (December 1998) contained one such mistake. The formula for total voltage in a parallel circuit, shown on page 40, is incorrect. The correct formula should read:

ET = ER1 = ER2 = ER3 = ER4 = ER5, etc.

Not So Simple!

When it comes to resistance in parallel, what seems so logical at first glance is simply not. The concept of resistance in parallel can best be understood and calculated by using mathematical formulas.

Take the circuit shown in Figure 1 as an example. This simple parallel circuit has two resistors in parallel. We see two 12 ohm resistors in the circuit. How much total resistance (RT) does the circuit have? At first glance it appears there are 24 ohms of resistance, which is the RT when these resistors are connected in series. But that is not the case when they are connected in parallel. In fact the total resistance (RT) of the two resistors in parallel is less than 12 ohms. How can two resistors in parallel have less resistance than either one alone? That's the understanding we're after here.

Let's try it first in words. Suppose a circuit includes one load (a bulb, a motor or whatever), a fuse and a switch all connected in a series across battery potential. Of course, this is a series circuit. Assuming nothing but the load has any resistance (an artificial simplification, of course), we can calculate from Ohm's Law that a 12 volt battery will push one amp through this circuit --voltage equals current times resistance, so current equals voltage divided by resistance.

In any series circuit, the total current (IT) exactly equals the current through any component, including the load (IR1). We know we have 12 volts in most automotive circuits, and we know the resistance of

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February 1999

this load is 12 ohms, so Ohm's Law says the current through R1 is one amp. As in every series circuit, all the current flows equally through each component in the circuit--through every connector, wire and the switch.

Parallel Complications

Now let's make a parallel circuit by adding a second load, R2, as shown in Figure 1. By adding R2 between the same two points connecting R1 to the circuit and ground, we've made the circuit parallel. Now electricity can flow in both of two parallel paths. Loads R1 and R2 are now connected in parallel. If R1 draws one amp and R2 is equal in resistance to R1, then R2 must draw exactly one amp, too. The battery must deliver two amps to the parallel circuit including R1 and R2. Ohm's Law still applies, of course, so we can still use RT = ET ? IT to calculate the actual total resistance of the parallel circuit. The battery is still 12 volts, but the total current has increased to two amps (IR1 + IR2) after the second load, R2, was added to the circuit in parallel.

If RT = ET ? IT RT = 12V ? 2A

RT = 6

According to Ohm's Law, R1 and R2 in parallel have a total resistance (RT) of 6 ohms, not 24 ohms (which might seem to be the total resistance at first glance). Ohm's Law doesn't have variations in assembly tolerances. Here's a simple rule for resistance in parallel which we see clearly illustrated in the above example:

The total resistance (RT) of a parallel circuit is always lower than the lowest individual resistance of any branch of the circuit.

In Figure 1, we have 12 ohms as the lowest resistance value in parallel, so we would know before doing any calculations that the total resistance must be less than 12 ohms, and 6 ohms is less than 12 ohms. So the general rule of total resistance in parallel is true. If you ever get a number for RT that is larger than the smallest individual resistance in a parallel circuit, you can bet you made a mistake in your calculations.

Practice! Practice! Practice!

1. If ET is 14.25 volts and IT is 3 amps, what is the RT of a parallel circuit with several resistors in parallel?

2. If ET is 13.49 volts and IT is 4.85 amps, what is the RT of a parallel circuit with several resistors in parallel?

3. How much resistance does a vehicle's computer memory have when the key is OFF and the keyoff drain is 12 mA, if ET is 12.66 volts? (Multiple computer memories are all connected in parallel with the voltage source. We'll assume all battery drain is through the computer).

4. How much resistance does a vehicle's computer memory have when the key is OFF and the keyoff drain is 75 mA, if ET is 12.66 volts?

5. Let's see how you do with a little more mental exercise in using information from the last few "Circuit Savvy" articles. Suppose the vehicle in Question 4 was parked for an extended time until the battery voltage (ET) dropped to 12.15 volts. What would the key-off drain current be with a lower ET? Note: We're calculating current, not resistance.

Calculating RT Without ET And IT

We can also determine RT without having to determine the values of ET and IT, which can save a little time. But you must know three formulas for resistance in parallel and use a basic hand-held calculator.

Formula #1

When all resistances in parallel are of equal value, as shown in Figure 2, the total resistance is found by the formula RT ? N. This formula is illustrated in Figure 3. Use the calculator to solve these problems.

B+

B+

Voltage

R1

Source

12

B-

R2 12

Figure 1: Two resistors in a parallel circuit.

B+ Voltage Source

B-

R1

R2

R3

R4

100

100

100

100

Figure 2: Four resistors in a parallel circuit.

February 1999

39

For any number of resistors in parallel with the same ohm value

(resistance of any one resistor)

R = RN T

(number of resistors)

Figure 3

RT = R ? N RT = 100 ? 4

RT = 25

Follow these step-by-step instructions to use a calculator to solve the above problem:

PRESS AC (ALL CLEAR) to clear all calculator functions and memory, PRESS 100 display shows 100 for 100 ohms, PRESS ? display shows 100, PRESS 4 display shows 4 for number of resistors in parallel, PRESS = display shows 25 (ohms), the answer.

Practice! Practice! Practice!

6. Four 36 ohm resistors are in parallel. What is their total resistance?

7. Three 3.2 ohm resistors are in parallel. What is their total resistance?

8. Three 16 ohm injectors are connected in parallel. What is their total resistance?

Formula #2

When two resistances with different ohm values are in parallel, the total resistance can be found using the product over the sum formula, as shown in Figure 4.

For exactly two resistors in parallel with different ohm values

R= T

Product Sum

=

R1 R1

x ?

R2 R2

Figure 4

Look at the schematic from Figure 1 again and suppose R1 is 120 ohms and R2 is 16 ohms. Both resistors have different values. We'll use a calculator and the special function keys to complete the equation. First calculate the sum and store it in memory. Then divide the product by the sum to obtain the answer.

Calculate the Sum of the two and store in memory as follows:

PRESS AC (ALL CLEAR) to clear all calculator functions and memory, PRESS 120 display shows 120, PRESS + to ADD the next number, PRESS 16 displays shows 16, PRESS = display shows 136, the sum of the two, PRESS M+ to store 136 in memory (display shows 136) On some calculators a small M is displayed to alert you that a number is stored in memory.

Calculate the Product of the two and divide by the memory value:

PRESS 120 display shows 120, PRESS X to multiply by the next number, PRESS 16 display shows 16, PRESS = displays shows 1920, the product of the two.

Calculate the answer to the problem which is labeled RT

With 1920 still showing on the display, PRESS ? (to divide by the next number). PRESS MR (Memory Recall) this returns the sum from memory for the next number and the display shows 136. PRESS = display shows 14.117647 (ohms), the value of RT as the calculator does the division.

Notice that 14.117647 ohms, rounded off to 14.12 ohms is lower than the smallest individual resistor value of 16 ohms, following the basic rule about resistance in parallel.

Practice! Practice! Practice!

Remember to PRESS AC (ALL CLEAR) to clear all calculator functions and memory after each problem is solved.

9. What is RT if R1 is 60 ohms and R2 is 29 ohms? 10. What is RT if R1 is 16 ohms and R2 is 12 ohms? 11. What is RT if R1 is 86 ohms and R2 is 119 ohms? 12. What is RT if R1 is 24 ohms and R2 is 39 ohms? 13. What is RT if R1 is 105 ohms and R2 is 147 ohms?

In every case notice that the answer for RT is always lower than the smallest individual resistor in parallel.

Formula #3

When more than two resistances are in parallel and they have different ohmic values, the total

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February 1999

resistance is found by the Reciprocal Formula, as shown in Figure 5.

For any three or more resistors in parallel with different ohmic values

R=

1

T

1 R1

+

1 R2

+

1 R3

Figure 5

The Reciprocal Formula looks complicated but the hand-held calculator will do the work for you if the keys are pressed in the correct sequence. It is always good practice to the sequence a few times until the procedure becomes familiar. Figure 6 shows three different resistor values inserted into the formula. R1 is 47 ohms, R2 is 68 ohms and R3 is 7 ohms.

Figure 6

R=

1

T

1 47

+

1 68

+

1 7

First calculate the equivalent of each fraction and add to memory. Then divide 1 by the summed memory value for the answer. Here's how it is done:

PRESS AC (ALL CLEAR) to clear all calculator functions and memory after each problem is solved, PRESS 1, PRESS ?, PRESS 47, PRESS = the readout shows 0.0212765, PRESS M+ to add to memory.

The first fraction is now added to memory. Notice the small "M" illuminating on the calculator readout signifying a number has been stored in memory.

PRESS 1, PRESS ?, PRESS 68, PRESS = the readout shows 0.0147058, PRESS M+ to add to memory.

The second fraction is now added to memory. PRESS 1, PRESS ?, PRESS 7, PRESS = the readout shows 0.1428571.

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February 1999

PRESS M+ to add to memory. The third fraction is now added to memory.

At this point all three fractions have been added together in memory. Next calculate the final step for RT. PRESS 1, PRESS ?, PRESS MR and the summed number in memory is placed under the 1, PRESS = the readout shows 5.591609 or 5.6 ohms rounded off for RT.

Practice! Practice! Practice!

Remember to PRESS AC (ALL CLEAR) to clear all calculator functions and memory after each problem is solved.

14. If three resistors, 56 ohms, 16 ohms and 47 ohms are in parallel, what is RT?

15. If three resistors, 86 ohms, 33 ohms and 10 ohms are in parallel, what is RT?

16. If three resistors, 22 ohms, 80 ohms and 2.2 ohms are in parallel, what is RT?

17. If three resistors, 95 ohms, 86 ohms and 68 ohms are in parallel, what is RT?

18. If three resistors, 9 ohms, 1.1 ohms and 1000 ohms are in parallel, what is RT?

Parallel Conclusions

Electric current, people say, always takes the path of least resistance. True enough, but not the whole truth; the path of least resistance is not the only path it takes. Don't get blinded to a basic principle of parallel circuits.

Figure 1 shows two resistors in a parallel circuit. Each provides a path for battery current. If R1 has twice the resistance of R2, which one will carry the higher current? The answer, obviously, is R2 with its lower resistance. But if we believe the people who say current always takes the path of least resistance, does that mean no current will flow through R1 since the R2 path has lower resistance? Of course

not. Both resistor paths carry battery current, but the current through R2, with lower resistance, is double the current through R1.

Current will take the path of least resistance when and only when that path is a dead short to ground, not simply because it has the lower resistance, but because it pulls system voltage down to zero. If in Figure 1 we run a wire from the top of R1 directly to ground, current will flow from the battery positive directly to ground, completely bypassing R1, until something on the circuit becomes the fuse and lets the smoke out of the wires or until the battery depletes. If battery voltage is restored or if the short is removed, current again flows through the two resistors in inverse proportion to their resistance: the higher the resistance in a branch, the lower that branch's current; the lower the resistance in that branch, the higher its current. The current in each parallel branch is entirely independent of the other branches, depending only on the resistance through that path and the capacity of the battery to sustain the voltage output.

--By Vince Fischelli

Answers:

1. RT = 4.75 ohms 2. RT = 2.78 ohms 3. RT = 1055 ohms 4. RT = 168.8 ohms 5. IT = ET ? RT, IT = 12.15 V ? 168.8 W, IT = 0.072 amps or 72 mA. We use the

168.8 ohms calculated in Question 4 because computer memory resistance is a fixed resistance that does not change. 6. RT = 9 ohms 7. RT = 1.067 ohms 8. RT = 5.34 ohms 9. RT = 19.55 ohms or 19.6 ohms when rounded off 10. RT = 6.86 ohms when rounded off 11. RT = 49.9 ohms when rounded off 12. RT = 14.86 ohms when rounded off 13. RT = 61.25 ohms when rounded off 14. 9.8 ohms when rounded off 15. 7.04 ohms when rounded off 16. 1.95 ohms when rounded off 17. 27.13 ohms when rounded off 18. 0.98 ohms when rounded off

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