Basic Electrical and Electronics Engineering: Unit I: Electrical Circuits
Solved Example Problems [Kirchoff's Laws]
Electrical Circuits
It states that the algebraic sum of the current meeting at a junction (or) node is equal to zero.
KIRCHOFF'S LAWS
Example: 16
Find the current through 12 Ω resistor as shown in figure.
Incoming current at point A is
5 + 6 = 11 A
The current through 12 Ω resistor is 11 A.
Example: 17
Find the voltage across the resistor 'R' in figure.
Applying KVL in the above circuit
20 - 8 + 10 - V1 – 17 + 5 = 0
10 – V1 = 0
10 = V1
Voltage across the resistor R = 10 V
Example: 18
For the circuit shown in figure, determine the unknown voltage drop V1,
Applying KVL in the above circuit
-2 -1 - V1 - 3 - 5 + 30 = 0
-V1 + 19 = 0
19 = V1
V1 = 19 V
Unknown voltage drop V1
Example: 19
Determine the current through resistance R3 in the circuit shown in figure.
According to Kirchoff's current law
IT = I1 + I2 + 13
50 = 30 + 10 + I3
I3 = 50 - 40
I3 = 10 mA
Example: 20
Find the current I and the voltage across 30 Ω shown in the figure.
Voltage drop across each resistor as
V8 = 8I, V30 = 30I, V2 = 2I
Applying KVL
100 - 40 = 8I + 30I + 2I
60 = 40I
I = 60/40 = 1.5 A
Voltage across 30 Ω = V30 = 30 I = 30 x 1.5 = 45 V
Example: 21
For the resistive circuit shown in fig. Find (a) I1 (b) VS.
(a) Applying KCL at node 1,
I1 = 4 – 1 = 3A
(b) V1 = 3 × 1 = 3V
V12 = 1 × 1 = 1V
V2 = 3 -1= 2V
I2 = V2/1 = 2/1 = 2A
KCL at node 2
I 3 = 1- I2 = 1- 2 = -1 A
VS = V2 - ( 1 × I3 ) = 2 - (1×-1)
=2+1=3V
Example: 22
Using KCL, find V2 for the given figure and given values i1 = 4A , V3 = 3V, V4 = 8V.
Applying KCL at the node
i1 – 3 i2 + i2 – i3 – i4 = 0
i1 – 2 i2 – i3 – i4 = 0
Substituting the values
4 - 2i2 - 1 – 2 = 0
1 – 2 i2 = 0
1 = 2 i2
i2 = 1/2 = 0.5 A
V2 = i2 × R = 0.5 x 2 = 1V
Basic Electrical and Electronics Engineering: Unit I: Electrical Circuits : Tag: : Electrical Circuits - Solved Example Problems [Kirchoff's Laws]
Basic Electrical and Electronics Engineering: Unit I: Electrical Circuits
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