1. Chapter 2
Resistive circuit
SAFIZAN BINTI SHAARI PPK MIKROELEKTRONIK

2. BASIC LAWS Ohm`s Law Kirchhoff’s Law
Series resistors & voltage division
Parallel resistors & current division
Y - transformation

3. KIRCHHOFF`S LAW Gustav Robert Kirchhoff
Born: 12 March 1824 in Königsberg, Prussia
Died: 17 Oct 1887 in Berlin, Germany
Kirchhoff's laws, which he announced in 1845,
allowed calculation of currents, voltages and resistances
in electrical circuits with multiple loops,
extending the work of Ohm.

4. KIRCHHOFF`S LAW 1 & 2 - parallel 10V & 4 - parallel
Two or more elements are in series if they exclusively share a single node and consequently share the same current.
Two or more elements are in parallel if they are connected to the same two nodes and consequently have the same voltage across them.
1 & 2 - parallel
10V & 4 - parallel
5 in series with
(1 and 2  in parallel)

5. KIRCHHOFF`S CURRENT LAW
Kirchhoff’s current law (KCL) states that the algebraic sum of currents entering a node (or a closed boundary) is zero
Mathematically,

6. KIRCHHOFF`S CURRENT LAW
Total current in = Total current out
Therefore:
I1 + I3 + I4 = I2+ I5 or
I1 + I3 + I4 – I2 – I5 = 0

7. KIRCHHOFF`S CURRENT LAW

8. KIRCHHOFF`S VOLTAGE LAW
Kirchhoff’s Voltage Law (KVL)
Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages around a closed path (or loop) is zero.
We can apply LOOP = CLOCKWISE or ANTI-CLOCKWISE
sum of voltage drop = sum of voltage rises
Mathematically,

9. SERIES RESISTORS &VOLTAGE DIVISION
Series: Two or more elements are in series if they are connected sequentially and consequently carry the same current.
The equivalent resistance of any number of resistors connected in a series is the sum of the individual resistances

10. SERIES RESISTORS &VOLTAGE DIVISION
Example:
A series circuit is one that
only one current path. R1 R1 R2 VS R2 VS R1 R2 R3 VS R3 R3 10 10

11. Example: Series circuit rule for current:
Because there is only one path, the current everywhere is
the same.
2.0 mA
For example, the reading on the first ammeter is 2.0 mA, What do the other meters read?
2.0 mA
2.0 mA
2.0 mA 11 11

12. SERIES VOLTAGE
VOLTAGE (V)
When two or more voltage sources are in series, the total voltage is equal to the the algebraic sum (including polarities of the sources) of the individual source voltages. 12 12

13. SERIES CIRCUIT Example: Voltage sources in series
Voltage sources in series add algebraically. For example, the total voltage of the sources shown is
27 V
What is the total voltage if one battery is reversed?
Question:
9 V 13 13

14. SERIES CIRCUIT Example:
For example, the resistors in a series circuit are 680 W, 1.5 kW, and 2.2 kW. What is the total resistance?
4.38 kW 14 14

15. SERIES RESISTORS &VOLTAGE DIVISION1 R 2 V _ + I -

16. SERIES RESISTORS &VOLTAGE DIVISION
Voltage at resistor R2:

17. VOLTAGE DIVISION Example: Series Resistance, R= 15 + 20 + 13 = 48 Ω
Rab = R48||16
= 16(48)/(16+48) = 12 Ω 17 17

18. VOLTAGE DIVISION Example:
Voltage dividers can be set up for a variable output using a potentiometer. In the circuit shown, the output voltage is variable.
Question:
What is the largest output voltage available?
5.0 V 18 18

19. PARALLEL RESISTORS &CURRENT DIVISION
Parallel: Two or more elements are in parallel if they are connected to the same two nodes and consequently have the same voltage across them.
The equivalent resistance of a circuit with N resistors in parallel is:

20. PARALLEL RESISTORS &CURRENT DIVISION

21. PARALLEL RESISTORS &CURRENT DIVISION

22. PARALLEL RESISTORS &CURRENT DIVISION

23. CURRENT DIVISION Example Current Divider
I1 = 4 Ω/(2 Ω + 4 Ω) × 30A = 20A
I2= 2 Ω /(2 Ω + 4 Ω) × 30A = 10A 23

24. Example 1: Calculating CurrentS1 R
1.8 kW B1
36 V V R =
36 V
1800 W
I =
= A
= 20 mA

25. Example 2: Calculating ResistanceB1
24 V A
0.03 A V I =
24 V
0.03 A
R =
= 800 W
= 0.8 k W

26. Example 3: Calculating VoltageR
270 W B1 A
0.15 A
V =
IR =
0.15 A x 270 W =
40.5 V

27. Example 4: Calculating PowerV
0.2 A
54 V
270 W
P =
IV =
0.2 A x 54 V =
10.8 W
P = I2R =
0.2 A x 0.2 A x 270 W =
10.8 W
P = V2/R =
(54 V x 54 V) / 270 W =
10.8 W

28. ANY QUESTIONS??