02.1 Resistors in Parallel Practice Problems

 

 

RESISTORS IN PARALLEL | PRACTICE PROBLEMS  

Complete the following problems to reinforce your understanding of the concept covered in this module.  

PROBLEM 1: Calculate the following for the circuit below using the principles of resistors in parallel. A. Equivalent resistance of the circuit B. Total current from the power supply, C. Current through each resistor D. Voltage drop across each resistor

     

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PROBLEM 2:

In the circuit shown below, two resistors are connected in parallel, R 1 = 2.2 Ω and R 2 = 1.5 Ω . Calculate the current through each resistor. Combine the resistors in parallel, and use Ohm’s law to obtain the total current from the battery.

   

PROBLEM 3:

For the circuit shown below find the equivalent resistance of the circuit shown below.

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RESISTORS IN PARALLEL | SOLUTIONS  

SOLUTION 1:

A. We can calculate the equivalent resistance of the circuit by combining the three resistors that in parallel using the formula for combining resistors in parallel:  

1 1 1 1 = + + R T R1 R2 R3 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 1 1 1 = + + R T 20Ω 100Ω 50Ω 1 = 0.05Ω + 0.01Ω + 0.02Ω RT

RT =

1 (0.05Ω + 0.01Ω + 0.02Ω)

R T = 12.5Ω

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  B. Since we have the voltage and resistance, we can calculate the total current from the power supply using Ohm’s Law:

V = I x R or R=

IT =

V V or I= I R

VT 125V = = 10A R T 12.5Ω

The formula for OHM’S LAW can be referenced under the topic of RESITIVITY on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. C. The current in each branch can be found using Ohm's law. OHM’S LAW:

V = I x R or R=

V V or I= I R

I1 =

V1 125V = = 6.25A R 1 20Ω

I2 =

V2 125V = = 1.25A R 2 100Ω

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I3 =

V3 R3

=

125V = 2.50A 50Ω

The formula for OHM’S LAW can be referenced under the topic of RESITIVITY on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. D. On a parallel circuit, each branch experiences the same voltage drop.

VT = V1 = V2 = V3 = 125V 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.

SOLUTION 2:

The resistors in parallel can be combined using the product-over-sum-rule:

RT =

R 1R 2 R1 + R2

The formula for the PRODUCT-OVER-THE-SUM-RULE can be referenced under the 14|PREPINEER.COM     psst…don’t  forget  to  take  notesÎ    

 

  topic of RESISTORS IN SERIES AND PARALLEL on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Plugging in the given resistance values:

RT =

2.2kΩ x 1.5kΩ = 0.892kΩ = 892Ω 3.7kΩ

R T = 892Ω Using Ohm’s Law we can calculate the current through each resistor and for the entire circuit:

V = I x R or R=

V V or I= I R

The formula for OHM’S LAW can be referenced under the topic of RESITIVITY on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Since the resistors are in parallel, the voltage is the same. Knowing this, we can calculate each current:

I1 =

5.0V = 2.27mA 2.2kΩ

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I2 =

5.0V = 3.33mA 1.5kΩ  

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  The total current from the battery is the total current flowing through the circuit. As we have calculated the equivalent resistance of the circuit, we can use Ohm’s Law to find the current:

IS =

5.0V = 5.60mA 892Ω

*We could also find this solution by summing I1 and I2.

SOLUTION 3:

In this problem we are looking for the equivalent resistance of the all the resistors in the circuit. To do this we identify groups of resistors that can be combined. The 8Ω and

7Ω resistors are in series and can be added together:

R 1 = 8Ω + 7Ω = 15Ω The calculated equivalent resistance R 1 and 10Ω are in parallel:

R2 =

1 1 1 + 10 R 1

=

10(R 1 ) = 6Ω 10 + R 1

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 16|PREPINEER.COM    

 

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  Reference Handbook, 9.3 Version for Computer Based Testing.

The

4Ω resistor and R 2 calculated equivalent resistance are in series. Their sum will

give us the equivalent resistance of the circuit:

R EQ = 4 + R 2 = 10Ω

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