# Basic Laws of Electronic Circuit

Topics: Series and parallel circuits, Kirchhoff's circuit laws, Electrical resistance Pages: 6 (699 words) Published: March 21, 2012
IED12102

Basic Laws:
1. 2. 3. 4. 5.

Electronic terminology Series Circuits Parallel Circuits Ohm’s Law and dc Circuits Simple Electrical Diagrams

For single-source parallel networks, the source current (I ) is equal to the sum of the individual branch currents. s

Is = I1 + I 2
For a parallel circuit, source current equals the sum of the branch currents. For a series circuit, the applied voltage equals the sum of the voltage drops.

V

V

In this first example, we will calculate the amount of current (I) in a circuit, given values of voltage (E) and resistance (R):

What is the amount of current (I) in this circuit?

15

In this second example, we will calculate the amount of resistance (R) in a circuit, given values of voltage (E) and current (I):

What is the amount of resistance (R) offered by the lamp?

16

In the last example, we will calculate the amount of voltage supplied by a battery, given values of current (I) and resistance (R):

What is the amount of voltage provided by the battery?

17

What if R=∞?
i=0 The Rest of the Circuit + v –

i = v/R = 0

What if R=0?
i The Rest of the Circuit + v=0 –

v=Ri=0

Kirchhoff’s Current Law (KCL)

sum of all currents entering a node is zero sum of currents entering node is equal to sum of currents leaving node Kirchhoff’s Voltage Law (KVL)

sum of voltages around any loop in a circuit is zero

A loop is any closed path through a circuit in which no node is encountered more than once Voltage Polarity Convention A voltage encountered + to - is positive A voltage encountered - to + is negative

I

Example 1: Determine current and source Since R3 and R4 are in parallel

V1 E V3 V2 R3 16Ω

R1 8Ω R2 6Ω I =3A 4 R4 8Ω

V3 = I 4 R4 = 3 × 8 = 24V = I 3 R3 = I 3 × 16
Therefore

24 I3 = = 1.5 A 16

By Kirchoff’s first law

Also

I = I 3 + I 4 = 1.5 + 3 = 4.5 A
By Kirchoff’s second law

V1 = IR1 = 4.5 × 8 = 36 V

V2 = IR2 = 4.5 × 6 = 27 V

E = V1 + V2 + V3 = 36 + 27 + 24 = 87 V

Example2: Determine I1, E, I3 and I

I V1 27V E V2

I1 R1 9Ω V3 R2 15Ω

V1 27 I1 = = = 3A R1 9

I3

R3 8Ω

V2 = I1 R2 = 3 ×15 = 45V
By Kirchoff’s second law (KVL)

E = V = V1 + V2 = 27 + 45 = 72V V 72 Also I3 = = = 9A R3 8
By Kirchoff’s first law (KCL)

I = I1 + I 3 = 3 + 9 = 12 A

EXAMPLE 3: Calculate the current in each branch First create loop current ,i.e I1 , I2, I3 as shown 20Ω 30Ω 40Ω

20Ω 3 30Ω I3 40Ω 50Ω 1 I1 20V 2 I2 50V 10Ω

60Ω

50Ω

10Ω

60Ω

100V

20V

50V

100V

Solving these equations

I1 = 1.65 A
Current in 60Ω Current in 30Ω Current in 50Ω Current in 40Ω Current in 10Ω Current in 20Ω

I 2 = 2.16 A
= I1 = 1.65 A
= I1 − I 3 = 0.15 A

I 3 = 1.50 A
In direction of I1 In direction of I1 In direction of I2 In direction of I2 In direction of I2 In direction of I3

= I 2 − I1 = 0.51A
= I 2 − I 3 = 0.66 A

= I 2 = 2.16 A
= I 3 = 1.50 A

100 − 20 = I1 (60 + 30 + 50) − I 2 50 − I 3 30 80 = 140 I1 − 50 I 2 − 30 I 3 ---(a)

In loop 1

50 + 20 = I 2 (50 + 40 + 10) − I1 50 − I 3 40
70 = −50 I1 + 100 I 2 − 40 I 3 ---(b)
0 = I 3 (30 + 20 + 40) − I1 30 − I 2 4060Ω
In loop 3
30Ω 1

In loop 2

20Ω 3 I3 40Ω 50Ω I1 20V 2 I2 50V 10Ω

0 = −30 I1 − 40 I 2 + 90 I 3

---(c)

100V

END