Chapter 05 Series Circuits Source: C-C Tsai Circuit Analysis: Theory and Practice Delmar Cengage Learning Series Circuits Two elements in a series Connected at a single point No other current-carrying connections at this point A series circuit is constructed by connecting various elements in series Current will leave the positive terminal of a voltage source, Move through the resistors, and return to negative terminal of the source C-C Tsai 2 1 Series Circuits Current is similar to water flowing through a pipe Current leaving the element must be the same as the current entering the element Same current passes through every element of a series circuit The laws, theorems, and rules that you apply to DC or AC circuits 3 C-C Tsai The Voltage Polarity Example 1 C-C Tsai Example 2 4 2 Kirchhoff’s Voltage Law (KVL) The algebraic sum of the voltage that rises and drops around a closed loop is equal to zero E 0 or ET - V1 - V2 - V3 - ∙∙∙ - Vn = 0 Another way of stating KVL is: Summation of voltage rises is equal to the summation of voltage drops around a closed loop ET = V1 + V2 + V3 + ∙∙∙ + Vn For example: E - V1 - V2 - V3 = 0 5 C-C Tsai Examples of Verifying KVL Example 1 C-C Tsai Example 2 6 3 Resistors in Series Most complicated circuits can be simplified For a series circuit V1 + V2 + V3 = E IR1 + IR2 + IR3 = E I (R1 + R2 + R3 )= E I (R1 + R2 + R3 )= I RT Total resistance RT in a series circuit is the sum of all the resistor values RT = R1 + R2 + R3 7 C-C Tsai Examples of Resistors in Series Example 1 RT =? C-C Tsai Example 2 RT =? 8 4 Power in a Series Circuit Power dissipated by each resistor is determined by the power formulas: P = VI = V 2/R = I 2R Since energy must be conserved, power delivered by voltage source is equal to total power dissipated by resistors PT = P1 + P2 + P3 + ∙∙∙ + Pn 9 C-C Tsai Examples of Power in a Series Circuit RT =? , I =? , and Voltage across each resistor Power dissipated by each resistor Power delivered by the voltage source Verify that total power dissipated by resistors is equal to power delivered by the voltage source Example 1 C-C Tsai Example 2 10 5 Voltage Sources in Series In a circuit with more than one source in series Example 1 Sources can be replaced by a single source having a value that is the sum or difference of the individual sources Polarities must be taken into account Resultant source Sum of the rises in one direction minus the sum of the voltages in the opposite direction C-C Tsai 11 Example: Voltage Sources in Series Example 2 C-C Tsai 12 6 Interchanging Series Components Order of series components May be changed without affecting operation of circuit Sources may be interchanged, but their polarities cannot be reversed After circuits have been redrawn, it may become easier to visualize circuit operation C-C Tsai 13 Example: Interchanging Series Components C-C Tsai 14 7 The Voltage Divider Rule Voltage applied to a series circuit Will be dropped across all the resistors in proportion to the magnitude of the individual resistors Rx Vx E RT Example 1 Voltage across each resistor 15 C-C Tsai Examples: Voltage Divider Rule Example 2 Voltage across each resistor C-C Tsai Example 3 Voltage across each resistor 16 8 Voltage Divider Rule Application If a single resistor is very large compared to the other series resistors, the voltage across that resistor will be the source voltage If the resistor is very small, the voltage across it will be essentially zero Example 4 Voltage across each resistor Example 5 Voltage across each resistor 17 C-C Tsai Circuit Ground Ground Point of reference or a common point in a circuit for making measurements One type of grounding is chassis ground In this type of grounding C-C Tsai Common point of circuit is often the metal chassis of the piece of equipment 18 9 Circuit Ground Chassis ground Earth ground Often connected to Earth Ground Physically connected to the earth by a metal pipe or rod If a fault occurs within a circuit, the current is redirected to the earth Voltages are often measured with respect to ground 19 C-C Tsai Voltage Subscripts For the circuit shown, we can express the voltage between any two node points (a and b) as Vab . If a is at a higher potential than b, then Vab is positive If b is at a higher potential than a, then Vab is negative Vab = Va – Vb Vba = Vb - Va Vab = -Vba C-C Tsai 20 10 Example: Double Subscripts Be sure to take the sign of the polarity on the same side of the source or element as you go around the circuit Vab = -50V Vbd =? Vac =? 21 C-C Tsai Example: Double Subscripts Vbd = ? Vea = ? C-C Tsai 22 11 Example: Double Subscripts (2) Vab = ? (1) Vab = ? 23 C-C Tsai Single Subscripts In a circuit with a ground reference point Voltages may be expressed with respect to that reference point Voltage Vb at point b with respect to ground Ground represents a point of zero reference potential Vb = ? C-C Tsai 24 12 Single Subscripts If voltages at various points in a circuit are known with respect to ground, then the voltage between points is easily determined Vab = Va – Vb = ? 25 C-C Tsai Point Sources Voltage source given with respect to ground May be represented by a voltage at a single point (node) in the circuit This voltage may be referred to as a point source Voltages at these points represent voltages with respect to ground, even if ground is not shown I=? C-C Tsai 26 13 Internal Resistance of Voltage Sources Ideal sources have no internal resistance In an ideal source Terminal voltage does not change when the load changes For a practical source, there is internal resistance 27 C-C Tsai Internal Resistance of Voltage Sources As the internal resistance changes Drop across the load changes Vab= 10.0V C-C Tsai Vab= 0.012V 28 14 Ammeter Loading Effects An ammeter is placed in a circuit to make a measurement Ammeter resistance will affect the circuit Amount of loading depends on instrument and circuit If the resistance of Ammeter is small compared to the resistance of the circuit, the loading effect will be small 29 C-C Tsai Ammeter Loading Effects Case1: RA (250 Ω) << R1 (20kΩ) I1= 0.5mA Measure I1 = 0.494mA Load effect = (0.5 - 0.494) / 0.5 = 1.23% C-C Tsai 30 15 Ammeter Loading Effects Case2: RA (250 Ω) > R1 (100Ω) I2= 0.5mA Measure I2 = 0.143mA Load effect = (0.5 - 0.143) / 0.5 = 71.4% 31 C-C Tsai Circuit Analysis using Multisim C-C Tsai 32 16 Kernel abilities 1. What is total resistance Rt of resistors R1~Rn in series? Please give an example. 2. What is KVL? Please give an example. 3. What is voltage divider? Please give an example. 4. What is loading effect of measuring current? Please give an example. 33 C-C Tsai Problem 7 C-C Tsai Solve unknown voltages 34 17 Problem 22 C-C Tsai Solve unknown voltage source 35 18