ELECTRIC AL CIRCUIT - An electric circuit is a continuous path along which an electric current can flow A simple circuit is composed of: 1. Power Source 2. Load 3. Conductors MAIN TYPES OF CIRCUIT 1. Closed Circuit – it is an uninterrupted path that allows a continuous flow of current through an electrical circuit. The switch completes the conductor path. To complete the circuit, a second conductor runs from the lamp back to the power source. 2. Open Circuit - the path of current flow is interrupted. The switch breaks the circuit. 3. Short Circuit - occurs when current leaks out of the intended conductor path such as out of a wire with damaged insulation. 4. Series Circuit - is connected so that current passes through each component in the circuit without branching off to individual components in the circuit. 5. Parallel Circuit - current branches off to individual components in the circuit KIRCHHOFF’S LAW 1st Law: Kirchhoff’s Voltage Law - Also known as Loop Rule. - states that “in any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop” which is also equal to zero. - States the relationship between voltage drops and voltage sources in a complete circuit. Assume that Voltage Drops are negative And Voltage Gains are positive ππππππ − ππππππ = 0 ππππππ = ππππππ Note: Kirchhoff’s Loop Rule was based on Law of Conservation of Energy KIRCHHOFF’S LAW 2nd Law: Kirchhoff’s Current Law - states that the “total current or charge entering a junction or node is exactly equal to the charge leaving the node as it has no other place to go except to leave, as no charge is lost within the node“ Assume that current going into the node are positive And current going out are negative πΌππ − πΌππ’π‘ = 0 πΌππ = πΌππ’π‘ Note: Kirchhoff’s Current Law was based on Law of Conservation of Charge SERIES AND PARALLEL CIRCUITS SERIES CIRCUIT PRINCIPLE πΌπππ‘ππ = πΌ1 = πΌ2 = πΌ3 = β― = πΌπ ππππ‘ππ = π1 + π2 + π3 + β― + ππ π πππ‘ππ = π 1 + π 2 + π 3 + β― + π π PARALLEL CIRCUIT PRINCIPLE πΌπππ‘ππ = πΌ1 + πΌ2 + πΌ3 + β― + πΌπ ππππ‘ππ = π1 = π2 = π3 = β― = ππ 1 π πππ‘ππ 1 1 1 = π + π +…+π 1 2 3 EXAMPLES 1. Six lamps are wired in a circuit. Each lamp draws 0.75 A and has a resistance of 110 Ω. Neglecting the effects of the wiring, determine the following: a) Total amperage in a series circuit b)Total amperage in a parallel circuit c) Total resistance in a series circuit d) Total resistance in a parallel circuit Solution: a) πΌπππ‘ππ = πΌ1 = πΌ2 = πΌ3 = β― = πΌ6 b) πΌπππ‘ππ = πΌ1 + πΌ2 + πΌ3 + β― + πΌ6 πΌπππ‘ππ = 0.75 π΄ π₯ 6 πΌπππ‘ππ = 4.5 π΄ c) π πππ‘ππ = π 1 + π 2 + π 3 + β― + π 6 π πππ‘ππ = 110 β¦ π₯ 6 π πππ‘ππ = 660 β¦ πΌπππ‘ππ = 0.75 π΄ d) π 1 πππ‘ππ 1 1 1 = π + π +…+π 1 π πππ‘ππ 1 = 2 6 110 β¦ 6 π πππ‘ππ = 18.33 β¦ 3. Determine the total current in the circuit. πΌ= π π 9π πΌ= 3 000 β¦ πΌ = 0.003 π΄ 4. Determine the total current in the circuit. ππ = 9 π π π = 3 πβ¦ + 10 πβ¦ + 5 πβ¦ π π = 18 πβ¦ πΌπ = ππ π π πΌπ = 9π 18 000 β¦ πΌπ = 0.002 π΄ 4. Determine the current in the every resistor πΌπ = 0.9 ππ΄ + 4.5 ππ΄ + 9 ππ΄ πΌπ = 14.4 ππ΄ ππ π π = πΌπ π π = V R1 9V I 0.9 mA R 10 kβ¦ R2 9V R3 9V 4.5 mA 9 mA 2 kβ¦ 1 kβ¦ Total 9V 14.4 mA Using Ohm’s Law Formula: πΌ1 = 9π 10 πβ¦ πΌ1 = 0.9 ππ΄ 9π πΌ2 = 2 πβ¦ 9π πΌ3 = 1 πβ¦ πΌ2 = 4.5 ππ΄ πΌ3 = 9 ππ΄ 9π 14.4 ππ΄ π π = 0.625 πβ¦ DIRECT & ALTERNATING CURRENT • Direct current (DC) is current flow in one direction in an electrical circuit. It is always from the negative to the positive terminals of the power source • Alternating current (AC) is a continuous reversal of the direction of current flow such that at a point in time the current flow is in one direction and at another point in time it is in the reverse direction. TRANSFORMING VOLTAGE AND CURRENT • A transformer is an electrical device that transfers an alternating current and voltage from one circuit to another using the induction phenomenon. The device is used in a circuit to change voltage, current, phase, and other electrical characteristics. • Transformer keeps the total power the same when voltage goes up or down. One must keep in mind that when the voltage goes up, the current goes down: π = πΌ1 π1 = πΌ2 π2 • The fundamental principle that allows transformers to change the voltage of alternating current is the direct relationship between the ratio of loops of wire in the primary winding to the secondary winding and the ratio of the primary voltage to the output voltage. The ratio between the number of turns (or loops) in the primary coil to the number of turns in the secondary coil is known as the turns ratio. The turns ratio establishes the following relationship with voltage: EXAMPLE • A 225 kVA transformer located outside a building is used to step down the voltage for the building. It is connected to a 7200 V AC power source. The ratio of the number of primary windings to the number of secondary windings on the transformer is 30 to 1. A) Approximate the voltage provided by the building. B) Approximate the current availbale in the building. Given: ππ = 7200 π ππ = 30 ππ = 1 Solution: ππ ππ = ππ ππ 30 7200 = 1 ππ ππ = 240 π π=ππ₯πΌ 225, 000= 240 π₯ πΌ πΌ = 937.5 π΄ IMPEDANCE & POWER FACTOR • Inductor - is a coil of wire that creates an electromagnetic field. • Capacitor - is composed of metal plates separated by air or a di-electric material such as paper, ceramic, or mica. Capacitors store electrical energy in an electrostatic field and release it later • Impedance (Z) is a measure of resistance to current flow on an AC circuit due to the combined effect of resistance, inductance and capacitance. Impedance is measured in ohms (Ω). Ohm’s Law for AC Circuit is: π = πΌπ EXAMPLE • The impedance values of incandescent lamps with three wattage ratings are noted below. With a voltage of 120 V on an AC circuit, determine the current of each lamp a) 100 W (121 Ω) b) 75 W (161 Ω) c) 40 W (300 Ω) Solution: π = πΌπ a) πΌ = π π 120 πΌ= 121 πΌ = 0.992 π΄ 120 b) πΌ = 161 πΌ = 0.745 π΄ 120 c) πΌ = 300 πΌ = 0.400 π΄ POWER FACTOR • Real power is the “working power” that performs useful effort in a circuit (e.g., creating heat, light, and motion); it is expressed in watts (W) or kilowatts (kW). • Reactive power is the power that generates the magnetic field required for inductive devices to operate. It dissipates no energy in the load but which returns to the source on each alternating current cycle; it is expressed in units called volt-amps-reactive (VAR) or kilovolt-amperes-reactive (kVAR), rather than watts. • The apparent power is the “power available to use.” It is expressed in volt-amperes (VA) or kilovolt-ampere (kVA), because it is the simple product of voltage and current. Apparent power is the “total” power required by an inductive device that is a composite (vector sum) of the real power and reactive power • The power factor (PF or cosφ) for a single-phase circuit is the ratio between real power and apparent power in a circuit: PF = real power/apparent power PF = W/VA EXAMPLES • A circuit consumes 3000 W of real power when the apparent power is 3600 VA. a) Determine the power factor. b) Determine phase angle φ. Solution: a) ππππ πππ€ππ ππΉ = ππππππππ‘ πππ€ππ 3000 π ππΉ = 3600 ππ΄ ππΉ = 0.833 b) cosΦ = 0.833 Φ = 33.6 πππ • An AC circuit is powering an electric heater (i.e., pure resistance, PF = 1.0). Assume the voltage is 240 V and a current draw of 10 A. Compute the apparent power and real power. Solution: ππ = ππ₯ πΌ ππ = ππΉ π₯ ππ΄ ππ = 240 π₯ 10 ππ = 2400 π₯ 1.0 ππ = 2400 π ππ = 2400 π • An AC circuit is powering a motor (i.e., inductive load, PF <1.0). Assume the voltage is 240 V and a current draw of 10 A. Compute the real power, assuming a power factor of 0.833.. ππ = ππ₯ πΌ ππ = ππΉ π₯ ππ΄ ππ = 240 π₯ 10 ππ = 2400 π₯ 0.833 ππ = 2400 π ππ = 1999 π
