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Chapter 5 & 6. Dr. Farid Farahmand CET 236. Outline. Identify a series/parallel circuit Determine the current and voltage in a circuit Determine total resistance Apply Ohm’s law Apply Kirchhoff’s voltage law. Series Circuit.
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Chapter 5 & 6 Dr. Farid Farahmand CET 236
Outline • Identify a series/parallel circuit • Determine the current and voltage in a circuit • Determine total resistance • Apply Ohm’s law • Apply Kirchhoff’s voltage law
Series Circuit • A series circuit provides only one path for current between two points • current is the same through each series resistor. • Find the total resistance in a series circuit • sum of the resistances of each individual series resistor. RT = R1 + R2 + R3 + . . . + Rn
Ohm’s Law in Series Circuits • If you know the total current, you can find the voltage drop across any of the series resistors by using: VR = ITR • The polarity of a voltage drop across a resistor is positive at the end of the resistor that is closest to the positive terminal of the voltage source • An open in a series circuit prevents current; and, there is zero voltage drop across each series resistor. • The total voltage appears across the points between which there is an open + -
Voltage Sources in Series • 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.
Kirchhoff’s Voltage Law (KVL) • The sum of all the voltage drops around a single closed loop in a circuit is equal to the total source voltage in that loop. VS = V1 + V2 + V3 + … + Vn Or VS - V1 - V2 - V3 = 0 • Since each resistor has the same current, the voltage drops are proportional to the resistance values. Vx = (Rx/RT)VS
Power in a Resistive Circuit • The total amount of power in a series/parallel resistive circuit is equal to the sum of the powers in each resistor in series. PT = P1 + P2 + P3 + . . . + Pn Remember Px = Vx.Ix • The amount of power in a resistor is important • the power rating of the resistor must be high enough to handle the expected power in the circuit.
Open and Short Circuit • When an open occurs in a series circuit, all of the source voltage appears across the open. • The most common failure in a series circuit is an open. • When a short occurs a portion of the series resistance is bypassed, thus reducing the total resistance. • A short in a series circuit results in more current than normal. What is the current when there is no short? What is the current when the circuit is shorted?
Open and Short Circuit • When an open occurs in a series circuit, all of the source voltage appears across the open. • The most common failure in a series circuit is an open. • When a short occurs a portion of the series resistance is bypassed, thus reducing the total resistance. • A short in a series circuit results in more current than normal.
Two Branches Resistors in Parallel • Each current path is called a branch. • A parallel circuit is one that has more than one branch. • The voltage across any given branch of a parallel circuit is equal to the voltage across each of the other branches in parallel
Kirchhoff’s Current Law (KCL) • Two ways of stating it: • The sum of the currents into a junction (total current in) is equal to the sum of the currents out of that junction (total current out). • The algebraic sum of all the currents entering and leaving a junction is equal to zero. IIN(1) + IIN(2) + . . . + IIN(n) = IOUT(1) + IOUT(2) + . . . +IOUT(m)
Total Parallel Resistance • When resistors are connected in parallel, the total resistance of the circuit decreases. • The total resistance of a parallel circuit is always less than the value of the smallest resistor. 1/RT = 1/R1 + 1/R2 + 1/R3 + . . . + 1/Rn or RT = R1||R2||R3||R4||R5….Rn What is the total resistance in a parallel circuit with 2 resistors? RT = R1R2/(R1 + R2)
Current in Parallel Circuits • The total current produced by all current sources is equal to the algebraic sum of the individual current sources. • A parallel circuit acts as a current divider because the current entering the junction of parallel branches “divides” up into several individual branch currents. I1=Vs/R1 = IT.RT/R1 I1 = (R2/(R1 + R2))IT I2 = (R1/(R1 + R2))IT Given IT, find I1 and I2!