04.1 KCL and KVL Practice Problems
KVL AND KCL | PRACTICE PROBLEMS
Complete the following problems to reinforce your understanding of the concept covered in this module.
PROBLEM 1:
What is the voltage across the 5 ! resistor in the center leg in the follow circuit?
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PROBLEM 2: Find the voltage v across the
4Ω resistor by the loop current method.
PROBLEM 3:
Find the unknown current in the circuit below
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KVL AND KCL | SOLUTIONS SOLUTION 1:
In this problem we will use Kirchoff’s Current Law and Kirchoff’s Voltage Law to find the current flowing through the circuit. Visually look at the circuit we see that is has 3 branches, and 2 nodes (A and B). We can then solve for the number of independent loops in the circuit using the formula relating branches, loops, and nodes in a circuit.
#LOOPS = #BRANCHES − #NODES + 1 Note: this formula is not provided in the reference manual
#LOOPS = 3 − 2 + 1 = 2 LOOPS
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We will use Kirchoff’s Voltage Law to derive equation in each loop in terms of the voltage entering and leaving the node. Then we will rewrite the equations is terms of current and resistance using Ohm’s Law. 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, 9.3 Version for Computer Based Testing. 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. Loop 1 is given as:
10V − V1 − V3 = 0 10V - R 1 × I1 + R 3 × (I2 + I3 )=0
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10 V − I1 5 Ω − I1 + I2 5 Ω = 0
Loop 2 is given as:
30V − V2 − V3 = 0 30V - R 2 × I2 + R 3 × (I2 + I3 )=0 14|PREPINEER.COM
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30 V − I2 5 Ω − I1 + I2 5 Ω = 0
Using the equation found for loop 2, we will solve for current I1 in terms of I2.:
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30 V − I2 5 Ω − I1 + I2 5 Ω = 0
I1 = 6V − 2I2
Now we plug this value into the first equation to solve for I2 :
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10 V − I1 5 Ω − I1 + I2 5 Ω = 0
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10 V − (6V − 2I2 ) 5 Ω − 6V − 2I2 + I2 5 Ω = 0
I2 =3.33 Amps
Substituting the calculated value of I2 into equation 21, we can solve for I1 :
30 V − I2 ( 5 Ω ) − ( I1 + I2 )( 5 Ω ) = 0 30 V − (3.33Amps) ( 5 Ω ) − ( I1 + 3.33Amps )( 5 Ω ) = 0 I1 =-0.667 Amps
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We can now use Ohm’s Law to calculate the voltage across the resistor in the center branch of the circuit:
V = IR V=(3.33 Amps - 0.667 Amps)(5 !)
V=13.33 V 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.
SOLUTION 2:
In this problem we will use Kirchoff’s Current Law and Kirchoff’s Voltage Law to find current flowing through the circuit. Visually look at the circuit we see that is has 3 branches, and 2 nodes (A and B). We can then solve for the number of independent loops:
#LOOPS = #BRANCHES − #NODES + 1 Note: this formula is not provided in the reference manual
#LOOPS = 3 − 2 + 1 = 2 LOOPS 16|PREPINEER.COM
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To find V in this circuit we assume current directions for the chosen loops and write Kirchoff’s voltage law for each loop as shown below: We will use Kirchoff’s Voltage Law to derive equation in each loop in terms of the voltage entering and leaving the node. Then, we will rewrite the equations in terms of current and resistance using Ohm’s Law. 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, 9.3 Version for Computer Based Testing. 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. Loop 1 is given as:
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6I1 +5I1 -5I2 +4 Volts=0 11I1 -5I2 =-4 Volts Loop 2 is given as:
-4 Volts+5I2 -5I1 -6 Volts+4I2 =0 -5I1 +9I2 =+10 Volts
Using the equation found for loop 2, we will solve for current, I1 in terms of I2 . 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.
9 I1 = −2V + I2 5
We will now plug this value into the first equation to solve for I2 :
11I1 -5I2 =-4 Volts
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9 11(-2V+ I2 )-5I2 =-4 Volts 5
I2 =1.216 Amps
Substituting the calculated value of I2 into equation 2, we can solve for I1 :
-5I1 +9I2 =+10 Volts
-5I1 +(9)(1.216 Amps)=+10 Volts I1 =0.19 Amps
We can now use Ohm’s Law to calculate the voltage across the
4Ω resistor.
V = IR V=(1.216 Amps)(4 !)
V=4.864 V
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.
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SOLUTION 3:
This problem is simply requiring you to use Kirchhoff’s Current Law and solve for the missing current value. 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, 9.3 Version for Computer Based Testing. Since we know that the sum of the currents entering and exiting the node must equal zero, we can simply set up an equation and solve for the missing current value.
2.6+4.4-3.3-I-1.7=0 I= 2 Amps
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