# Kirchhoff's Laws

• Kirchhoff’s Current Law (KCL)

• Kirchhoff’s Voltage Law (KVL)

#### Kirchhoff’s Current Law (KCL):

 The algebraic sum of all currents entering a node must always be zero $\sum_{n-1}^{N}i_{n}=0$ where in is the nth current. N is the number of branches. A common assignment: if the current is entering the node, assign a negative “-“sign and if the current is leaving the node, assign a positive “+” sign. For the following figure The node equation can be written as $-i_{1}+i_{2}-i_{3}+i_{4}=0$ To use KCL to analyze a circuit, Write KCL equations for the currents Use Ohm’s law to write currents in terms of bode voltages (one equation for each resistor) Solve to find values of node voltage and current

Example: Find the current through a 20Ω resistance, and current through a 40Ω resistance

 Write KCL at node x $-i_1+i_2-2A=0$ Write $v_x$ in the circuit using Ohm’s Law $i_1=\frac{12V-v_x}{20\Omega},$    $i_2=\frac{v_x}{40\Omega}$ Apply last two equation into KCL at node x $-i_1+i_2-2A=-\frac{12V-v_x}{20\Omega}+\frac{v_x}{40\Omega}-2A=0$ $-0.6+0.05v_x+0.025v_x-2=0$ $v_x=34.67V$ The current through a 20Ω resistance $i_1=\frac{12V-34.67}{20\Omega}=-1.133A$ The current through a 40Ω resistance $i_2=\frac{v_x}{40\Omega}=\frac{34.67}{40\Omega}=0.866A$

#### Kirchhoff’s Voltage Law (KVL):

The algebraic sum of all voltage around the closed loop must be always zero.

 $\sum_{n-1}^{N}v_{n}=0$ where vn is the nth voltage. N is the number of elements in the loop A common assignment: if the positive (+) side of the voltage is encountered first, assign a positive “+”sign to the voltage across the element. If the negative (-) side of the voltage is encountered first, assign a negative “-”sign to the voltage across the element. For the following figure $\sum_{n=1}^{N}V_{n}=-12+V_{1}+V_{2}-6+V_{3}+V_{4}=0$ $V_{1}+V_{2}+V_{3}+V_{4}=12+6$ $V_{1}+V_{2}+V_{3}+V_{4}=18$ To use KVL to analyze a circuit, Write KVL equations for voltages Use Ohm’s law to write voltages in terms of resistances and currents. Solve to find values of the currents and then voltages.

### Examples:

Example 2: Find the current i and voltage v over the each resistor.

 KVL equations for voltages $v_{1}+v_{2}+v_{3}+v_{4}=18(1)$ Using Ohm’s Law $v_{1}=10\Omega,$ $v_{2}=20\Omega,$ $v_{3}=40\Omega,$ $v_{4}=20\Omega$ Substituting into KVL equation $10i+20i+40i+20i=18$ $(90)i=18$ $i=\frac{18}{90}=0.2A$ $v_{1}=R_{1}i=10(0.2)=2V,$     $v_{2}=R_{2}i=20(0.2)=4V$ $v_{3}=R_{3}i=40(0.2)=8V,$     $v_{4}=R_{4}i=20(0.2)=-4V$

Example 3: Find v1 and v2 in the following circuit
(note: the arrows are signifying the positive position of the box and the negative is at the end of the box)

 Loop 1 $-V_1+1+4=0$ $V_1=5V$ Loop 2 $-4+3+V_2=0$ $V_2=1V$

Example 4: Find V1, V2, and V3.
(note: the arrows are signifying the positive position of the box and the negative is at the end of the box)

 Loop 1 $-20-25+10+V_1=0$ $V_1=35V$ Loop 2 $-10+15-V_2=0$ $V_2=5V$ Loop3 $-V_1+V_2+V_3=0$ $-35+5+V_3=0$ $V_3=30V$

Example 5: Find V1, V2, V3,and V4
(note: the arrows are signifying the positive position of the box and the negative is at the end of the box)

 Loop 1 $-V_4+2+5=0$ $V_4=7V$ Loop 2 $4+V_3+V_4=0$ $V_3=-4-7$ $V_3=-11V$ Loop 3 $-3+V_1-V_3=0$ $V_1=V_3+3$ $V_1=-8V$ Loop 4 $-V_1-V_2-2=0$ $V_2=-V_1-2$ $V_2=6V$

### Practice Problems:

(Click image to view solution)

Problem 1: Find V1 in the following circuit.

#### View Solution

 Solution: By KVL $-9+(6+2)i+3=0$ $8i=9-3=6\Rightarrow{i=\frac{6}{8}=0.75A}$ By KVL for inner loop $-9+6i+V_1=0$ $V_1=9-6i$ $V_1=9-6(0.75)=4.5V$

Problem 2: Find V0 in the following circuit.

#### View Solution

 Solution: KVL Outher Loop $-30-10+8+I(3+5)=0$ $8I=32$ $I=4A$ KVL right inner loop $-V_0+5I+8=0$ $V_0=8+5I=8+5(4)=28V$

Problem 3: Find V1,V2, and V3 in the following circuit.

#### View Solution

 Solution: Outher lopp $-24+V_1+10+12=0$ $V_1=2V$ V2 and 10V source loop $V_2+10+12=0$ $V_2=-22V$ V3 and 10V source loop $-V_3+10=0$ $V_3=10V$

Problem 4: Find  I1,I2, I3in the following circuit

#### View Solution

 Solution: Loop 1 $-6=-2I_1+3((I_2-I_1)$ Loop 2 $12=3(I_2-I_1)+4(I_2-I_3)$ Loop 3 $24=-4(I_3-I_2)$ From Loop 3 $(I_2-I_3)=6$ Substitute into loop  2 , the results is $12=3(I_2-I_1)+4(6)$ $I_2-I_1=-4$ (4) Substitute into loop  1, the results is $-6=-2I_1+3(-4)$ $I_1=-3A$ Substitute into l(4), the results is $I_2-(-3A)=-4A$ $I_2=-7A$ Substitute into loop  3, the results is $24=-4(I_3+7)$ $-6=I_3+7$ $I_3=-13A$

Problem 5: Find the resistor R value in the following circuit.

#### View Solution

 Solution: 200 ohms and 50 Ohms resistors have same voltage and they are parallel $R_p=\frac{200\times50}{200+50}=40\Omega$ $I=\frac{10}{40}=0.25A$ The voltage on R resistor is $V_R=15-10=5V$ Then, we can find the resistor value as $R=\frac{V_R}{I}=\frac{5}{.25}=20\Omega$

### Exercises:

1. V1=8V, V2=-4V, V4=14V. Find V3 and V5 in the following circuit

2. Find Vx and Vy in the following circuit
3. Find Vx , Vy and Vz in the following circuit
4. Find the KCL node equations at nodes A,B,C, and D

5. If I1=4A , I2 =5A, and I3 =3A, then using KCL find I4 and, I5 in the following circuit
1. V3 =12V and V5=-2V
2. Vx =12V and Vy=9V
3. Vx =35V, Vy=5V, and Vz=15V
4. At node A:
$I_1+I_2=0$
At node B:
$-I_2+I_3+I_4=0$
At node C:
$-I_4+I_5+I_6=0$
At node D:
$-I_1-I_3-I_5-I_6=0$
5. I4=2A and I5=1A