Kirchhoff's Current & Voltage Laws:
Kirchhoff's Voltage Law-KVL: The algebraic sum of the voltages around any closed path is zero.
Network: It is a portion of an electric circuit, contains some circuit elements. Means that each circuit is a network, but not necessarily that a network be a circuit.
Network: It is a portion of an electric circuit, contains some circuit elements. Means that each circuit is a network, but not necessarily that a network be a circuit.
It means that if we trace out a closed path, the algebraic sum of the voltages across the individual elements around it must be zero. This may be written in the form of equation:
v1 + v2 + v3 + · · · + vN = 0
Or, in compact form:
Σv = 0
Σv = 0
Tracing a closed path is performed either a clockwise or a counter-clockwise direction. In both cases, we assume that visiting a circuit element from its +ve terminal adds a +ve voltage to the equation, whereas visiting a circuit element from its -ve terminal results in adding a -ve voltage to the equation.
Example (1): Consider the following circuit. Find both vx and ix.
Solution:
It is required that we find both ix and vx, but to get the value of ix by applying Ohm's law, you have to know the value of vx which can be obtained using KVL.
Applying KVL in clockwise direction, we get:
− 5 − 7 + vx = 0
Solving for vx, we get:
vx = 12V
Now, apply Ohm's law to get the value of ix:
ix = vx/Rx
ix = 12/100
ix = 0.12A = 120mA
*************************
Example (2): For the following circuit, find both ix and vx.
Solution:
It is similar to the example above.
You can also apply KVL to get the value of ix instead of vx, this way the value of vx in the equation should be substituted with ix * Rx according to Ohm's law.
Applying KVL in clockwise direction, we get:
+ 3 + 1 + vx = 0
But vx = ix * Rx
+ 3 + 1 + ix * Rx = 0
+ 3 + 1 + 10ix = 0
ix = - 0.4A = - 400mA
Using Ohm's law, the value of vx is easily obtained:
vx = ix * Rx
vx = - 0.4 * 10 = - 4V
*************************
Example (3): Consider the following circuit which contains 8 circuit elements. Find vR2 and vx.
Solution:
You can assume these bubbles as any circuit element (We don't care about what they are really), we are just interesting about the voltage across these bubbles.
As it is required from you to find vR2, you should look for a loop which results in only one unknown in case of applying KVL. This loop is the left-most loop (with respect to R2). Applying KVL in clockwise direction we get:
+ 4 - 36 + vR2 = 0
vR2 = 32V
You also need to find the value of vx. This case if you want to apply KVL taking all circuit elements in consideration within the loop you selected (either the big loop or the right-most loop) you will encounter three unknowns, which cannot be solved within one equation, but you are given that vx is the voltage across the network consisting v2, vs1 & vR1 (The voltage across network a-b is = - v2 - vs1 + vR1 = vx). So, instead of having three unknowns, you can have vx in the equation:
Applying KVL to the big loop, we get:
+ 4 - 36 + 12 + 14 + vx = 0
vx = 6V
Surly, you are not forced to use this solution, you can choose any loop which satisfies your requirements.
*************************
Example (4): For The following circuit, find both vR2 and v2, if vR1 = 1V.
Solution:
It is the same idea of the last example, you have the network a-b again.
You need to find the value of vR2 which can be determined by applying KVL around the small left-most loop, this results in:
+ 8 - 12 + vR2 = 0
Note that the polarity of the resistor is determined only by the direction you visit is with, so the voltage of the resistor is assumed always +ve even if it is mentioned -ve in the question (Always consider it +ve).
Solving for vR2, we get:
vR2 = 4V
To get the value of v2, vx must be determined.Applying KVL to the big loop we get:
+ 8 - 12 + 7 - 9 + vx = 0
Solving for vx, we have:
vx = 6V
But vx = - v2 - 3 + vR1
and vR1 is given 1V in question. So:
6 = - v2 - 3 + 1
- v2 = 8V
v2 = - 8V
*************************
Example (5): Consider the following circuit and find vx.
Solution:
You cannot use KVL directly to find vx because the current in the 4Ω resistor is different from the current in the 8Ω resistor for example. So, this requires finding the values of the currents flowing at nodes, that means you will need both KCL & KVL.
Solution:
It is required that we find both ix and vx, but to get the value of ix by applying Ohm's law, you have to know the value of vx which can be obtained using KVL.
Applying KVL in clockwise direction, we get:
− 5 − 7 + vx = 0
Solving for vx, we get:
vx = 12V
Now, apply Ohm's law to get the value of ix:
ix = vx/Rx
ix = 12/100
ix = 0.12A = 120mA
*************************
Example (2): For the following circuit, find both ix and vx.
Solution:
It is similar to the example above.
You can also apply KVL to get the value of ix instead of vx, this way the value of vx in the equation should be substituted with ix * Rx according to Ohm's law.
Applying KVL in clockwise direction, we get:
+ 3 + 1 + vx = 0
But vx = ix * Rx
+ 3 + 1 + ix * Rx = 0
+ 3 + 1 + 10ix = 0
ix = - 0.4A = - 400mA
Using Ohm's law, the value of vx is easily obtained:
vx = ix * Rx
vx = - 0.4 * 10 = - 4V
*************************
Example (3): Consider the following circuit which contains 8 circuit elements. Find vR2 and vx.
Solution:
You can assume these bubbles as any circuit element (We don't care about what they are really), we are just interesting about the voltage across these bubbles.
As it is required from you to find vR2, you should look for a loop which results in only one unknown in case of applying KVL. This loop is the left-most loop (with respect to R2). Applying KVL in clockwise direction we get:
+ 4 - 36 + vR2 = 0
vR2 = 32V
You also need to find the value of vx. This case if you want to apply KVL taking all circuit elements in consideration within the loop you selected (either the big loop or the right-most loop) you will encounter three unknowns, which cannot be solved within one equation, but you are given that vx is the voltage across the network consisting v2, vs1 & vR1 (The voltage across network a-b is = - v2 - vs1 + vR1 = vx). So, instead of having three unknowns, you can have vx in the equation:
Applying KVL to the big loop, we get:
+ 4 - 36 + 12 + 14 + vx = 0
vx = 6V
Surly, you are not forced to use this solution, you can choose any loop which satisfies your requirements.
*************************
Example (4): For The following circuit, find both vR2 and v2, if vR1 = 1V.
Solution:
It is the same idea of the last example, you have the network a-b again.
You need to find the value of vR2 which can be determined by applying KVL around the small left-most loop, this results in:
+ 8 - 12 + vR2 = 0
Note that the polarity of the resistor is determined only by the direction you visit is with, so the voltage of the resistor is assumed always +ve even if it is mentioned -ve in the question (Always consider it +ve).
Solving for vR2, we get:
vR2 = 4V
To get the value of v2, vx must be determined.Applying KVL to the big loop we get:
+ 8 - 12 + 7 - 9 + vx = 0
Solving for vx, we have:
vx = 6V
But vx = - v2 - 3 + vR1
and vR1 is given 1V in question. So:
6 = - v2 - 3 + 1
- v2 = 8V
v2 = - 8V
*************************
Example (5): Consider the following circuit and find vx.
Solution:
You cannot use KVL directly to find vx because the current in the 4Ω resistor is different from the current in the 8Ω resistor for example. So, this requires finding the values of the currents flowing at nodes, that means you will need both KCL & KVL.
You will need to assign directions for the currents flowing in the circuit, this is done by your assumption. Don't worry, if the value of the current is -ve, this means that your assumed direction is just opposite to the real flow direction. the following figure shows the circuit with the assumed current directions:
Applying KVL at the left-most loop (The place where you can have less unknowns), we get:
- 60 + 8 * 5 + v10 = 0
v10 = 20V
By Ohm's law, we can get the value of i10:
i10 = 20/10
i10 = 2A
Apply KCL at the upper left node, we get:
5 = 2 + i4
i4 = 3A
Now we can find the value of vx by applying KVL at the big loop, we get:
- 60 + 5 * 8 + 3 * 4 + vx = 0
vx = 8V
*************************
**************************************************- 60 + 8 * 5 + v10 = 0
v10 = 20V
By Ohm's law, we can get the value of i10:
i10 = 20/10
i10 = 2A
Apply KCL at the upper left node, we get:
5 = 2 + i4
i4 = 3A
Now we can find the value of vx by applying KVL at the big loop, we get:
- 60 + 5 * 8 + 3 * 4 + vx = 0
vx = 8V
*************************
Reference: Engineering Circuit Analysis 8th Edition
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