Chapter 2 Resistive circuit SAFIZAN BINTI SHAARI PPK MIKROELEKTRONIK.

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Presentation transcript:

Chapter 2 Resistive circuit SAFIZAN BINTI SHAARI PPK MIKROELEKTRONIK

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

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.

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)

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,

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

KIRCHHOFF`S CURRENT LAW

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,

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

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

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

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

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

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

SERIES RESISTORS &VOLTAGE DIVISION 1 R 2 V _ + I -

SERIES RESISTORS &VOLTAGE DIVISION Voltage at resistor R2:

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

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

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:

PARALLEL RESISTORS &CURRENT DIVISION

PARALLEL RESISTORS &CURRENT DIVISION

PARALLEL RESISTORS &CURRENT DIVISION

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

Example 1: Calculating Current S1 R 1.8 kW B1 36 V V R = 36 V 1800 W I = = 0.02 A = 20 mA

Example 2: Calculating Resistance B1 24 V A 0.03 A V I = 24 V 0.03 A R = = 800 W = 0.8 k W

Example 3: Calculating Voltage R 270 W B1 A 0.15 A V = IR = 0.15 A x 270 W = 40.5 V

Example 4: Calculating Power V 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

ANY QUESTIONS??