04 KCL and KVL FINAL
KCL AND KVL | CONCEPT OVERVIEW
The topic of KIRCHOFF’S LAWS can be referenced on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. A NODE is a point where several wires are electrically connected and is usually symbolized by a dot or circle at the wire crossing. It may also be referred to as a JUNCTION. A BRANCH is a line with elements that connects two nodes. A LOOP is an independent closed path in a circuit. The relationship between nodes, branches, and loops is expressed:
#of BRANCHES = #of LOOPS + #of NODES − 1 Note: this formula is not provided in the reference handbook. KIRCHOFF’S CURRENT LAW (KCL) states that the “total current or charge entering a junction or node is exactly equal to the charge leaving the node as it has no other place to go except to leave, as no charge is lost within the node“. 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. 1|PREPINEER.COM
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In other words the algebraic sum of ALL the currents entering and leaving a node must be equal to zero:
IEXITING + IENTERING = 0
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Current Entering - Current Leaving = I1 +I2 +I3 + -I4 +-I5 =0
It is important to utilize a sign convention that you can easily remember as you work your way through circuit analysis problems. It is typical to assume that currents entering the node are positive, and currents leaving the node are negative. That is how we will address problems from here on out. KIRCHOFF’S VOLTAGE LAW states that in any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop, which is also equal to zero. In other words the algebraic sum of all voltages within the loop must be equal to zero. 2|PREPINEER.COM
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VAB +VBC +VCD +VDA =0 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.
PROCEDURES TO USE KIRCHOFF’S CIRCUIT LAWS: 1. Determine the number of loops using the following formula:
#LOOPS = #BRANCHES − #NODES + 1 Note: this formula is not provided in the reference manual
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2. Apply Kirchoff’s Current Law at each node to derive an equation for the current entering and leaving the node.
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. 3. Apply Kirchoff’s Voltage Law to each loop to derive an equation for the voltage entering and leaving the loop. If you are applying Kirchoff’s Voltage Law to one loop, and an element has two or more loops, the total voltage drop across the element equals the sum (or difference) of the voltage drops due to the individual loop currents.
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. 4. Convert the equations created in step 3 to be in terms of the equations solved for in step 2 or vice versa. This can be done using Ohm’s Law.
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. 5. Using the rules of simultaneous equations, and substitution, solve for the variables in each of the equations. Then, plug them into the other set of equations to solve for 4|PREPINEER.COM psst…don’t forget to take notesÎ
all of the variables. It is important to remember that the direction of a loop current may be either clockwise or counter-clockwise but once the choice is made do NOT change it.
Hint: Use the equation solve function on your calculator to avoid using having to do various algebra calculations and fraction manipulations by hand. This is helpful for saving time and avoiding simple algebra errors.
CONCEPT EXAMPLE: Find the current flowing across the 40 ! Resistor, R 3 .
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SOLUTION: In this problem we will use Kirchoff’s Current Law and Kirchoff’s Voltage Law to find the current in each loop of the circuit and the resistance of the
40Ω Resistor, R 3 .
Visually looking 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
#LOOPS = 3 − 2 + 1 = 2 LOOPS Note: this formula is not provided in the reference manual The next step is for us to apply Kirchoff’s Current Law at each node, and derive equations of the circuit entering and leaving the node. At node A:
I1 + I2 - I3 = 0 I1 + I2 = I3
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At node B:
I3 - I1 - I2 = 0 I3 = I1 + 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. Next we will use Kirchoff’s Voltage Law to derive an equation in each loop in terms of the voltage. We will then write the voltage in terms of the 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:
10 − V1 − V3 = 0 7|PREPINEER.COM
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10 = R 1 × I1 + R 3 × I3 = 10I1 + 40I3
10 = 10I1 + 40I3
Loop 2 is given as:
20 − V2 − V3 = 0 20 = R 2 × I2 + R 3 × I3 = 20I2 + 40I3
20 = 20I2 + 40I3 We will then treat the entire circuit as a loop since it is a closed loop. The previous two loops were independent loops. Loop 3 is given as:
10 – 20 = 10I1 – 20I2 As I3 is the sum of I1 + I2 we can rewrite I3 as:
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10 = 10I1 + 40 I1 + I2 = 50I1 + 40I2 20 = 20I2 + 40 I1 + I2 = 40I1 + 60I2 8|PREPINEER.COM
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We now have two “Simultaneous Equations” that can be reduced to give us the values of I1 and I2 Substitution of I1 in terms of I2 gives us the value of I1 as:
I1 = -0.143 Amps Substitution of I2 in terms of I1 gives us the value of I2 as:
I2 = +0.429 Amps Since we know I3 = I1 + I2
The current flowing in resistor R 3 is given as:
I3 = -0.143 + 0.429 = 0.286 Amps Using Ohm’s Law, the voltage across the resistor R 3 is given as:
V3 = 0.286 × 40 = 11.44 volts
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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, 9.3 Version for Computer Based Testing. The negative sign for I1 means that the direction of current flow initially chosen was wrong, but the absolute value is still correct.
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