Resistors in Parallel
Applying KCL to more complex circuits.
Kirchhoff’s Current Law Example
Circuit Resistance RAC
Circuit Resistance RCF
Thus the equivalent circuit resistance between nodes C and F is calculated as 10 Ohms. Then the total circuit current, IT is given as:
RT = R(AC) + R(CF) = 1 + 10 = 11Ω
Giving us an equivalent circuit of:
Therefore, V = 132V, RAC = 1Ω, RCF = 10Ω’s and IT = 12A.
Having established the equivalent parallel resistances and supply current, we can now calculate the individual branch currents and confirm using Kirchhoff’s junction rule as follows.
VAC = IT X RAC = 12 x 1 = 12 Volts
VCF = IT x RCF = 12 x 10 = 120 Volts
Thus, I1 = 5A, I2 = 7A, I3 = 2A, I4 = 6A, and I5 = 4A.
We can confirm that Kirchoff’s current law holds true around the circuit by using node C as our reference point to calculate the currents entering and leaving the junction as:
At node C ∑IIN = ∑IOUT
IT = I1 + I2 = I3 + I4 + I5
∴ 12 = (5 + 7) = (2 + 6 + 4)
We can also double check to see if Kirchhoff's Current Law holds true as the currents entering the junction are positive, while the ones leaving the junction are negative, thus the algebraic sum is: I1 + I2 – I3 – I4 – I5 = 0 which equals 5 + 7 – 2 – 6 – 4 = 0. So we can confirm by analysis that Kirchhoff’s current law (KCL) which states that the algebraic sum of the currents at a junction point in a circuit network is always zero is true and correct in this example.
Kirchhoff’s Current Law Example
Find the currents flowing around the following circuit using Kirchhoff’s Current Law only.
IT is the total current flowing around the circuit driven by the 12V supply voltage. At point A, I1 is equal to IT, thus there will be an I1*R voltage drop across resistor R1.
The circuit has 2 branches, 3 nodes (B, C and D) and 2 independent loops, thus the I*R voltage drops around the two loops will be:
Loop ABC ⇒ 12 = 4I1 + 6I2
Loop ABD ⇒ 12 = 4I1 + 12I3
Since Kirchhoff’s current law states that at node B, I1 = I2 + I3, we can therefore substitute current I1 for (I2 + I3) in both of the following loop equations and then simplify.
Kirchhoff’s Loop Equations
We now have two simultaneous equations that relate to the currents flowing around the circuit.
Eq. No 1 : 12 = 10I2 + 4I3
Eq. No 2 : 12 = 4I2 + 16I3
By multiplying the first equation (Loop ABC) by 4 and subtracting Loop ABD from Loop ABC, we can be reduced both equations to give us the values of I2 and I3
Eq. No 1 : 12 = 10I2 + 4I3 ( x4 ) ⇒ 48 = 40I2 + 16I3
Eq. No 2 : 12 = 4I2 + 16I3 ( x1 ) ⇒ 12 = 4I2 + 16I3
Eq. No 1 – Eq. No 2 ⇒ 36 = 36I2 + 0
Substitution of I2 in terms of I3 gives us the value of I2 as 1.0 Amps
Now we can do the same procedure to find the value of I3 by multiplying the first equation (Loop ABC) by 4 and the second equation (Loop ABD) by 10. Again by subtracting Loop ABC from Loop ABD, we can be reduced both equations to give us the values of I2 and I3
Eq. No 1 : 12 = 10I2 + 4I3 (x4) ⇒ 48 = 40I2 + 16I3
Eq. No 2 : 12 = 4I2 + 16I3 (x10) ⇒ 120 = 40I2 + 160I3
Eq. No 2 – Eq. No 1 ⇒ 72 = 0 + 144I3
Thus substitution of I3 in terms of I2 gives us the value of I3 as 0.5 Amps
As Kirchhoff’s junction rule states that : I1 = I2 + I3
The supply current flowing through resistor R1 is given as : 1.0 + 0.5 = 1.5 Amps
Thus I1 = IT = 1.5 Amps, I2 = 1.0 Amps and I3 = 0.5 Amps and from that information we could calculate the I*R voltage drops across the devices and at the various points (nodes) around the circuit.
68 videos|85 docs|62 tests
|
1. What is Kirchhoff's current law? |
2. How can Kirchhoff's current law be applied in practical circuits? |
3. Can Kirchhoff's current law be violated? |
4. How is Kirchhoff's current law related to Kirchhoff's voltage law? |
5. Can Kirchhoff's current law be applied to both DC and AC circuits? |
|
Explore Courses for Electrical Engineering (EE) exam
|