it's law of kircchoff's

1. ACTIVELEARNINGASSIGnMENT
CIRCUITS AND NETWORKS(2130901)
Prepared by
Group - 1
Div- B
Sem-3rd
Branch-Electrical
Guided by: Prof. Megha Ma’m
GANDHINAGAR INSTITUTE OF Technology.

2. Brijesh Patel -140120109030
Kavan Patel -140120109032
Darshil Shah -140120109050
Group : 1
PREPAREDBY:
ENROLLMENTNO.

3. TOPIC NAME:
•KIRCHHOFF’S LAWS.

4. TYPES OF LAWS:
Kirchhoff’s laws.
KIRCHHOFF’S CURRENT LAW(KCL) KIRCHHOFF’S VOLTAGE LAW (KVL)
These laws are first derived by Guatov Robert
Kirchhoff and hence these laws are also referred
as Kirchhoff Laws.

5. Circuit Definitions
 Node – any point where 2 or more circuit elements
are connected together
 Wires usually have negligible resistance
 Each node has one voltage (w.r.t. ground)
 Branch – a circuit element between two nodes
 Loop – a collection of branches that form a closed
path returning to the same node without going
through any other nodes or branches twice

6. Kirchhoff's Current Law
(KCL)
 The algebraic sum of currents entering a node is zero
 Add each branch current entering the node and subtract
each branch current leaving the node
 Σ currents in - Σ currents out = 0
 Or Σ currents in = Σ currents out
(IB + IC + ID) - IA = 0
Then, the sum of all the
currents is zero. This can be
generalized as follows,

7. Kirchhoff's Voltage Law
(KVL)
“The algebraic sum of all voltages around a
closed loop must be equal to zero”.
 Σ voltage drops - Σ voltage rises = 0
 Or Σ voltage drops = Σ voltage rises
That means, V1 + V2 + V3 + ................... + Vm −
Vm + 1 − Vm + 2 − Vm + 3 + .....................− Vn = 0.
So accordingly Kirchhoff Second
Law,
∑V = 0.

8. Ideal voltage and current source.
VOLTAGE SOURCE CURRENT SOURCE

9. Voltage rise and voltage
drop:
Voltage drop
Voltage rise

10. Example.
 Find the current through 8Ω resistor using KIRCHHOFF’S LAWS.

11. EXAMPLE OF KCL

12. EXAMPLE OF KCL
the nodal equations (We assume all currents are leaving the node):
0
7
v102
5
2
2
12
0
2
21
6
v51
1
v121

















k
V
k
V
k
VV
k
VV
k
V
k
V
Rearranging the terms:
Solving the linear equations with the Cramer’s Rule:
100259135
7723110


VV
VV
495.3
485
1695
2748.8
485
4243
1
1695
10035
7710
4243
59100
377
485
5935
310













VV

13. EXAMPLE OF KCL
Currents:
 I1k = = -3.252 mA () I2k = = 2.626 mA ()
 I3k = = 1.928 mA () I4k = = 1.928 mA ()
 I5k = = 0.699 mA () I6k = = 0.625 mA ()

14. EXAMPLE OF KVL

15. EXAMPLE OF KVL
 Junction J1:
I = I1 + I2 (equation 1)
 Junction J2:
I1 + I2 = I (equation 2)

16.  Loop A: (start from the upper left corner and move clockwise)
-I1 x (100 Ω) + 1.5V = 0 (equation 3)
Therefore: I1 = 0.015 A
 Loop B:
-9V – I2 x (200 Ω) + I1 x (100 Ω) = 0 (equation 4)
 Substituting the value of I1 into equation 3 yields:
-9 – I2 x (200 Ω) + (0.015)(100 Ω) = 0
-7.5 = (200) x I2 therefore: I2 = -0.0375 A
 And then I = -0.0225 A
 VR1 = I1 x R1 = (0.015 A) x (100 Ω) therefore VR1 = 1.5V
 VR2 = I2 x R2 = (-0.0375 A) x (200 Ω) therefore VR2 = -7.5V

17. REFERENCES.
www.electricals4u.com
www.eefundamentals.com
www.electricaleasy.com
www.wikipedia.com/KCL,KVL
CIRCUITS & NETWORKS book