Table of contents  
Introduction 
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Resistors in Parallel
Applying KCL to more complex circuits.
Kirchhoff’s Current Law Example
Circuit Resistance R_{AC}
Circuit Resistance R_{CF}
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, R_{AC} = 1Ω, R_{CF} = 10Ω’s and I_{T} = 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.
V_{AC} = I_{T} X R_{AC} = 12 x 1 = 12 Volts
V_{CF} = I_{T} x R_{CF} = 12 x 10 = 120 Volts
Thus, I_{1} = 5A, I_{2} = 7A, I_{3} = 2A, I_{4} = 6A, and I_{5} = 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 ∑I_{IN} = ∑I_{OUT}
I_{T} = I_{1} + I_{2} = I_{3} + I_{4} + I_{5}
∴ 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: I_{1} + I_{2} – I_{3} – I_{4} – I_{5} = 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.
I_{T} is the total current flowing around the circuit driven by the 12V supply voltage. At point A, I_{1} is equal to I_{T}, thus there will be an I_{1}*R voltage drop across resistor R_{1}.
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 = 4I_{1} + 6I_{2}
Loop ABD ⇒ 12 = 4I_{1} + 12I_{3}
Since Kirchhoff’s current law states that at node B, I_{1} = I_{2} + I_{3}, we can therefore substitute current I_{1} for (I_{2} + I_{3}) 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 : 1_{2} = 10I_{2} + 4I_{3}
Eq. No 2 : 1_{2} = 4I_{2} + 16I_{3}
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 I_{2} and I_{3}
Eq. No 1 : 12 = 10I_{2} + 4I_{3} ( x4 ) ⇒ 48 = 40I_{2} + 16I_{3}
Eq. No 2 : 12 = 4I_{2} + 16I_{3} ( x1 ) ⇒ 12 = 4I_{2} + 16I_{3}
Eq. No 1 – Eq. No 2 ⇒ 36 = 36I_{2} + 0
Substitution of I_{2} in terms of I_{3 }gives us the value of I_{2} as 1.0 Amps
Now we can do the same procedure to find the value of I_{3} 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 I_{2} and I_{3}
Eq. No 1 : 12 = 10I_{2} + 4I_{3} (x4) ⇒ 48 = 40I_{2} + 16I_{3}
Eq. No 2 : 12 = 4I_{2} + 16I_{3} (x10) ⇒ 120 = 40I_{2} + 160I_{3}
Eq. No 2 – Eq. No 1 ⇒ 72 = 0 + 144I_{3}
Thus substitution of I_{3} in terms of I_{2} gives us the value of I_{3} as 0.5 Amps
As Kirchhoff’s junction rule states that : I_{1 }= I_{2} + I_{3}
The supply current flowing through resistor R_{1} is given as : 1.0 + 0.5 = 1.5 Amps
Thus I_{1} = I_{T} = 1.5 Amps, I_{2} = 1.0 Amps and I_{3 }= 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.
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