Kirchhoff’s Circuit Laws Explained

A Simple Guide to Kirchhoff’s Laws

Learning the basic rules of electric circuits!

Kirchhoff’s First Law: The Current Law (KCL)

The Big Idea: All the electricity that flows into a point must flow out of it.

In simple terms: Nothing gets lost! The amount of electric current going in is always equal to the amount coming out.

Analogy: Imagine a busy crossroads. The number of cars driving into the crossroads must equal the number of cars driving out of it. Electricity (current) works the same way at any connection point (node) in a circuit!

Rule: Current In = Current Out

KCL Examples (20 Examples)

Example 1: Find Ix

2A3AIx

In: 2A

Out: 3A, Ix

2 = 3 + Ix

Ix = -1A (means 1A flows in)

Example 2: Find Ix

5A2AIx1A

In: 5A, 2A

Out: Ix, 1A

5 + 2 = Ix + 1

Ix = 6A

Example 3: Find Ia

10AIa4A3A

In: 10A, Ia, 4A

Out: 3A

10 + Ia + 4 = 3

Ia = -11A (11A flows out)

Example 4: Find I1

I17A12A

In: 12A

Out: I1, 7A

12 = I1 + 7

I1 = 5A

Example 5: Find Iy

6A2AIy

In: 6A

Out: 2A, Iy

6 = 2 + Iy

Iy = 4A

Example 6: Find Iz

15A8AIz

In: 15A

Out: 8A, Iz

15 = 8 + Iz

Iz = 7A

Example 7: All currents leaving

2A5AIx

In: 0

Out: 2A, 5A, Ix

0 = 2 + 5 + Ix

Ix = -7A (7A flows in)

Example 8: All currents entering

3A4AIx

In: 3A, 4A, Ix

Out: 0

3 + 4 + Ix = 0

Ix = -7A (7A flows out)

Example 9: Find Ib

1A2AIb5A

In: 2A

Out: 1A, Ib, 5A

2 = 1 + Ib + 5

Ib = -4A (4A flows in)

Example 10: Find Ic

8AIc8A

In: 8A, 8A

Out: Ic

8 + 8 = Ic

Ic = 16A

Example 11: Find Id

Id3A9A

In: Id, 9A

Out: 3A

Id + 9 = 3

Id = -6A (6A flows out)

Example 12: Find Ie

12AIe12A

In: 12A

Out: Ie, 12A

12 = Ie + 12

Ie = 0A

Example 13: Find If

2A3A4AIf

In: 2A, 3A

Out: 4A, If

2 + 3 = 4 + If

If = 1A

Example 14: Find Ig

Ig5A5A

In: Ig

Out: 5A, 5A

Ig = 5 + 5

Ig = 10A

Example 15: Find Ih

6A2AIh

In: 6A, 2A

Out: Ih

6 + 2 = Ih

Ih = 8A

Example 16: Find Ix

5AIx

In: 5A

Out: Ix

5 = Ix

Ix = 5A

Example 17: Find Ix

1AIx

In: 1A, Ix

Out: 0

1 + Ix = 0

Ix = -1A (1A flows out)

Example 18: Find Ix

Ix3A3A

In: Ix

Out: 3A, 3A

Ix = 3 + 3

Ix = 6A

Example 19: Find Ix

10A3AIx7A

In: 10A

Out: 3A, Ix, 7A

10 = 3 + Ix + 7

Ix = 0A

Example 20: Find Ix

4AIx9A

In: Ix, 9A

Out: 4A

Ix + 9 = 4

Ix = -5A (5A flows out)

Kirchhoff’s Second Law: The Voltage Law (KVL)

The Big Idea: The energy you start with in a loop is the energy you end with.

In simple terms: In any complete circle (loop) in a circuit, the energy given by the battery must be completely used up by all the parts in that loop (like light bulbs or resistors).

Analogy: Think about climbing up a playground slide and then sliding down. You start on the ground. The ladder (like a battery) gives you “energy” to go up. When you slide down (like going through a light bulb), you use up that energy. By the time you reach the bottom, you’re back where you started with zero extra height. In a circuit loop, all the energy given must be used up!

Rule: Energy Gained = Energy Used

KVL Examples (20 Examples)

Example 21: Find VR2

+12VR1V=5VR2V=?

Gained: 12V

Used: 5V + VR2

12V = 5V + VR2

VR2 = 7V

Example 22: Find VS

+VS=?R1=3VR2=6VR3=2V

Gained: VS

Used: 3V + 6V + 2V

VS = 11V

VS = 11V

Example 23: Find VR3

+24VR1=10VR2=8VR3=?

Gained: 24V

Used: 10V + 8V + VR3

24V = 18V + VR3

VR3 = 6V

Example 24: Find VR2

+20VR1=8V+5VR2=?

Loop: +20V – 8V + 5V – VR2 = 0

17V – VR2 = 0

VR2 = 17V

Example 25: Find VR1

+9VR1=?R2=4V

Gained: 9V

Used: VR1 + 4V

9V = VR1 + 4V

VR1 = 5V

Example 26: Find VR1

+10VR1=?+5VR2=12V

Loop: +10V – VR1 – 5V – 12V = 0

-7V – VR1 = 0

VR1 = -7V (polarity is backwards)

Example 27: Find VS2

+50VR1=20V+VS2=?R2=15V

Loop: +50V – 20V + VS2 – 15V = 0

15V + VS2 = 0

VS2 = -15V

Example 28: Find VR2

+18VR2=?R1=18V

Gained: 18V

Used: VR2 + 18V

18V = VR2 + 18V

VR2 = 0V

Example 29: Find VS

+VS=?R1=7VR2=3V

Gained: VS

Used: 7V + 3V

VS = 10V

VS = 10V

Example 30: Find Vx

+40VR1=15VR2=VxR3=15V

Gained: 40V

Used: 15V + Vx + 15V

40V = 30V + Vx

Vx = 10V

Example 31: Find VS

?VSR1=5VR2=5V

Gained: VS

Used: 5V + 5V

VS = 10V

VS = 10V

Example 32: Find VR3

+100VR1=25VR2=50VR3=?

Gained: 100V

Used: 25V + 50V + VR3

100V = 75V + VR3

VR3 = 25V

Example 33: Find VS1

+VS1=?R1=12V+3VR2=9V

Loop: +VS1 – 12V + 3V – 9V = 0

VS1 – 18V = 0

VS1 = 18V

Example 34: Find VR1

+15VR1=?+5VR2=5V

Loop: +15V – VR1 – 5V – 5V = 0

5V – VR1 = 0

VR1 = 5V

Example 35: Find VR2

+6VR1=9VR2=?

Loop: -6V – 9V – VR2 = 0

-15V – VR2 = 0

VR2 = -15V (polarity is backwards)

Example 36: Find VR2

+10VR1=2VR2=?

Gained: 10V

Used: 2V + VR2

10V = 2V + VR2

VR2 = 8V

Example 37: Find VS

+VS=?R1=5VR2=15V

Gained: VS

Used: 5V + 15V

VS = 20V

VS = 20V

Example 38: Find VR3

+30VR1=10VR2=10VR3=?

Gained: 30V

Used: 10V + 10V + VR3

30V = 20V + VR3

VR3 = 10V

Example 39: Find VR1

+12VR1=?+4V

Loop: +12V – VR1 + 4V = 0

16V – VR1 = 0

VR1 = 16V

Example 40: Find VS

+VS=?R1=8VR2=4V

Gained: VS

Used: 8V + 4V

VS = 12V

VS = 12V

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