01.1 Resistors in Series Practice Problems
RESISTORS IN SERIES | PRACTICE PROBLEMS Complete the following problems to reinforce your understanding of the concept covered in this module.
PROBLEM 1: Determine the following quantities for the circuit shown below: A. The equivalent resistance of the circuit B. The total current from the power supply C. The current flowing through each resistor D. The voltage drop across each resistor
PROBLEM 2:
Two resistors (R1= 2.2 kΩΩ and R2 = 1.5 kΩΩ) are connected in series. Calculate the resistor series equivalent. If the battery voltage is 5 V (constant), calculate the current. 8|PREPINEER.COM
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Using Ohm’s law, calculate the voltage across each resistor. Do the two resistor voltages add up to 5 V?
PROBLEM 3: A battery supplies 30 V to a circuit with R 1 = 10Ω and R 2 = 20Ω in series. What is the current in the circuit, and what is the voltage drop across each resistor?
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RESISTORS IN SERIES | SOLUTIONS SOLUTION 1: A. Calculate the equivalent resistance: The resistors are in series so the equivalent resistance is the sum of the resistors in the circuit:
R T = R1 + R2 + R3 R T = 20 Ω + 30 Ω + 50 Ω = 100 Ω The formula for RESISTORS IN SERIES can be referenced under the topic of RESISTORS IN SERIES AND PARALLEL on page 200 of the NCEES Supplied Reference Handbook, 9.2 Version for Computer Based Testing. B. Calculate the total current from the power supply: Given the voltage and the equivalent resistance we just calculated, we can use Ohm’s Law to calculate the total current of the circuit:
IT =
VT 125V = = 1.25A R T 100Ω
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The formula for OHM’S LAW can be referenced under the topic of RESITIVITY on page 200 of the NCEES Supplied Reference Handbook, 9.2 Version for Computer Based Testing. C. Calculate the current through each resistor: Remember, the current does not change in resistors that are in series. Therefore the current through each resistor is equal to the total current of the circuit:
I T = I1 = I2 = I3 = 1.25A The properties of RESISTORS IN SERIES can be referenced under the topic of RESISTORS IN SERIES AND PARALLEL on page 200 of the NCEES Supplied Reference Handbook, 9.2 Version for Computer Based Testing. D. Calculate the voltage drop across each resistor: The voltage drops can be found using Ohm's law.
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, 9.2 Version for Computer Based Testing. 11|PREPINEER.COM
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Using the given values of the resistors (R1= 20 ΩΩ, R2 = 30 ΩΩ, and R3=50ΩΩ) and the calculated current (I1=I2=25A), we can solve for the voltage drop at each resistor:
V1 = I1 x R 1 = (1.25A)(20Ω) = 25.0V V2 = I2 x R 2 = (1.25A)(30Ω) = 37.5V V3=I3 x R3 = (1.25A)(50ΩΩ)=62.5V
SOLUTION 2: In the problem we are given: • Resistance: R 1 = 2.2 kΩ R 2 = 1.5 kΩ • Voltage: V=5V The two resistors are in series, therefore the equivalent resistance of the circuit is equal to the sum of the two resistors.
RS = R1 + R2 Plugging in our resistance values:
R S = 2.2kΩ + 1.5kΩ = 3.7kΩ 12|PREPINEER.COM
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We can use Ohm’s Law to calculate the current as we given the voltage and resistance of the circuit:
V = I x R or R=
I=
V V or I= I R
5.0V = 1.351mA 3.7kΩ
The formula for OHM’S LAW can be referenced under the topic of RESITIVITY on page 200 of the NCEES Supplied Reference Handbook, 9.2 Version for Computer Based Testing. Since the resistors are in series, the current is constant through each resistor. We can plug in the calculated current and given resistance values to find the voltage at each resistor:
V1=R1 x I = (2.2kΩΩ)(1.35 mA) = 2.97 V V2=R2 x I = (1.5kΩΩ)(1.35 mA) = 2.03 V
V1 = 2.97V V2 = 2.03V
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According to Kirchhoff’s Voltage Law, the total voltage supplied is equal to the sum of the voltages across each resistor:
V = V1 + V2 The formula for KIRCHOFF’S VOLTAGE LAW can be referenced under the topic of KIRCHOFF’S LAWS on page 200 of the NCEES Supplied Reference Handbook, 9.2 Version for Computer Based Testing.
2.97V + 2.03V = 5.00V Since the sum of the voltages across each resistor equals the total voltage supplied to the circuit, our answer is correct.
SOLUTION 3: In the problem we are given: • Resistance: R 1 = 10 kΩ R 2 = 20 kΩ • Voltage: V=30 V As the resistors are in series, the equivalent resistance is the sum of the resistances of the two resistors:
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R EQ = R 1 + R 2 R EQ = 10Ω + 20Ω = 30Ω
Using Ohm’s Law we can solve for the current of the circuit:
I=
V 30V = = 1A R 30Ω
The formula for OHM’S LAW can be referenced under the topic of RESITIVITY on page 200 of the NCEES Supplied Reference Handbook, 9.2 Version for Computer Based Testing. Remembering that current (I) is constant, we know the resistance and current of each resistor., We can now use Ohm’s Law to calculate the voltage drop across each resistor:
V = IR V1 = IR 1 = (1A)(10Ω) = 10V V2 = IR 2 = (1A)(20Ω) = 20V To check our answer we sum the voltage across each resistor to make sure it equals the total voltage supplied to the circuit:
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V1 + V2 = V 10V + 20V = 30V
The formula for KIRCHOFF’S VOLTAGE LAW can be referenced under the topic of KIRCHOFF’S LAWS on page 200of the NCEES Supplied Reference Handbook, 9.2 Version for Computer Based Testing. Since the sum of the voltages across each resistor is equal to the voltage supplied to the circuit, our calculations for the voltage drop across each resistor is correct.
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