02 Resistors in Parallel FINAL

 

 

RESISTORS IN PARALLEL | CONCEPT OVERVIEW  

The topic of RESISITIVITY can be referenced on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.  

When multiple RESISTORS are in the same circuit, they are either considered in parallel, in series, or a combination of both. RESISTORS that are connected in PARALLEL have the same VOLTAGE DROP through each resistor. The properties of RESISTORS IN PARALLEL can be referenced under the topic of RESISTORS IN SERIES AND PARALLEL on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.

The EQUIVALENT RESISTANCE of PARALLEL RESISTORS is calculated by:

RP =

1 1 1 1 + + ... + R1 R2 Rn

The formula for RESISTORS IN PARALLEL can be referenced under the topic of RESISTORS IN SERIES AND PARALLEL on page 200 of the NCEES Supplied 1|PREPINEER.COM     psst…don’t  forget  to  take  notesÎ    

 

  Reference Handbook, 9.3 Version for Computer Based Testing. The sum of the CURRENT entering a node equals the sum of the currents leaving that node. This relationship is known as KIRCHOFF’S CURRENT LAW.

IT = I1 + I2 + ... + In The formula for KIRCHOFF’S CURRENT LAW can be referenced under the topic of KIRCHOFF’S LAWS on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. If there are only two resistors in parallel, the equivalent resistance can be determined by a PRODUCT-OVER-THE-SUM-RULE.

RT =

R 1R 2 R1 + R2

The formula for the PRODUCT-OVER-THE-SUM-RULE can be referenced under the topic of RESISTORS IN SERIES AND PARALLEL on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.  

CONCEPT EXAMPLE:

For the circuit shown below, there are multiple parallel resistors connected between the terminals a and b. Determine the equivalent resistance of the circuit. 2|PREPINEER.COM    

 

psst…don’t  forget  to  take  notesÎ  

 

 

SOLUTION:

For this type of problem the key is to stay organized and identify groups of resistors you can combine to find an equivalent resistance. The first step is to group resistors that are in parallel or in series so you can find the equivalent resistance of that group and simplify the circuit. The three resistors in the lower right corner are in parallel so we will use them as our starting point.  

  3|PREPINEER.COM    

 

psst…don’t  forget  to  take  notesÎ  

 

  Since we will have multiple equivalent resistance calculations we will use numeric notation to keep them organized:

1 R EQ1 1 R EQ1

=

1 1 1 + + 2Ω 2Ω 1Ω

= 3.0Ω

R EQ1 = 0.5Ω

The formula for RESISTORS IN PARALLEL can be referenced under the topic of RESISTORS IN SERIES AND PARALLEL on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. We can now replace the three resistors that were in parallel with the equivalent resistance R EQ1 = 0.5Ω

4|PREPINEER.COM    

 

psst…don’t  forget  to  take  notesÎ  

 

  Since R EQ1 is in series with the 2.0Ω resistor, we can combine them to find R EQ2

R EQ2 = 2.0Ω + R EQ1 = 2.0Ω + 0.5Ω R EQ2 = 2.5Ω We can now replace the two resistors with R EQ2

The next group of resistors we can combine is the 3.0 ! resistor and 5.0 Ω resistor at the top of the circuit that are connected in parallel. 5|PREPINEER.COM    

 

psst…don’t  forget  to  take  notesÎ  

 

 

1 R EQ3 1 R EQ3

=

1 1 1 + = (0.33 + 0.20) 3Ω 5Ω Ω

= 0.53

R EQ3 =

1 Ω

1 Ω 0.53

R EQ3 = 1.89Ω

The formula for RESISTORS IN PARALLEL can be referenced under the topic of RESISTORS IN SERIES AND PARALLEL on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. We can now replace the 3.0 Ω resistor and 5.0 Ω resistor that are connected in parallel with the equivalent resistance R EQ3 . 6|PREPINEER.COM    

 

psst…don’t  forget  to  take  notesÎ  

 

 

The equivalent resistance R EQ3 and the 1.0 Ω resistor are in series, so we can combine them to calculate the equivalent resistance R EQ4 .

R EQ4 = R EQ3 + 1Ω R EQ4 = 2.89Ω We can now replace the 3.0 Ω resistor and 5.0 Ω resistor that are connected in parallel with the equivalent resistance R EQ3 .

7|PREPINEER.COM    

 

psst…don’t  forget  to  take  notesÎ  

 

 

Since R EQ4 is in parallel with R EQ2 , we can combine them to find R EQ5 .

1 R EQ5

=

1 R EQ4

+

1 R EQ2

The formula for RESISTORS IN PARALLEL can be referenced under the topic of RESISTORS IN SERIES AND PARALLEL on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.

1 R EQ5

=

R EQ5 =

1 1 1 1 + = (0.346 + 0.4) = 0.746 2.89Ω 2.5Ω Ω Ω 1 Ω = 1.34Ω 0.746

8|PREPINEER.COM    

 

psst…don’t  forget  to  take  notesÎ  

 

We can now replace equivalent resistances R EQ4

  and R EQ2 , that are connected in

parallel with the equivalent resistance R EQ5 .

Since R EQ5 is in series with the 2.0Ω resistor, we can combine them to find R EQ for the total circuit.

R EQ = 2.0Ω + R EQ5 R EQ = 2.0Ω + 1.34Ω = 3.34Ω

R EQ = 3.34Ω

9|PREPINEER.COM    

 

psst…don’t  forget  to  take  notesÎ