Into -electric Circuit- Pres

Electrotechnology DC Circuit

DC Circuit

Specific Learning Objectives

 Apply circuit law to solve problem in dc networkseries, parallel, combined dc and parallel circuits  Solve problem to find power and efficiency in electromechanical system and consideration required when apply the laws in network problems  Solve problem regarding resistance required to extent range of ammeter and voltmeter and verify the experiments

Electric Circuit.

Electric Circuit

Electric Circuit

Electric Circuit

Electric Circuit

Electric Circuit

Electric Circuit

Electric Circuit

DC Circuit

DC Circuit

DC Circuit

Ohm’s law • states that the current I flowing in a circuit is directly proportional to the applied voltage V and inversely proportional to the resistance R, provided the temperature remains constant. Thus, Electrical power • Power P in an electrical circuit is given by the product of potential difference V and current I. The unit of power is the watt, W. Hence Electrical energy • Electrical energy = power × time • If the power is measured in watts and the time in seconds then the unit of energy is watt-seconds or joules. • If the power is measured in kilowatts and the time in hours then the unit of energy is kilowatt-hours, often called the ‘unit of electricity’. • The ‘electricity meter’ in the home records the number of kilowatt-hours used and is thus an energy meter.

DC Circuit

Main effects of electric current • The three main effects of an electric current are: • Magnetic effect : bells, relays, motors, generators, transformers, telephones, car-ignition and lifting magnets. • Chemical effect : primary and secondary cells and electroplating • Heating effect : cookers, water heaters, electric fires, irons, furnaces, kettles and soldering irons

DC Circuit

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Series circuits Figure 5.1 shows three resistors R1, R2 and R3 connected end to end, i.e., • in series, with a battery source of V volts.

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Since the circuit is closed a current I will flow and the p.d. across each resistor may be determined from the voltmeter readings V1, V2 and V3

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In a series circuit the current I is the same in all parts of the circuit and hence the same reading is found on each of the two ammeters shown. • the sum of the voltages V1, V2 and V3 is equal to the total applied voltage

from Ohm’s Law

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DC Circuit • •

Potential divider The voltage distribution for the circuit shown in Figure 2.2(a) is given by:Figure 2.2 : divider voltage - series • Source : Electrical and Electronic 8th Edition (2002) • The circuit shown in Figure 2.2(a) is often referred to as a potential divider circuit. Such a circuit can consist of a number of similar elements in series connected across a voltage source, voltages being taken from connections between the elements.

•Frequently the divider consists of two resistors as shown in Figure 2.2(a), where

DC Circuit

Parallel networks • Figure 2.3 shows three resistors, R1, R2 and R3 connected across each other, i.e., in parallel, across a battery source of V volts. In a parallel circuit: • a) the sum of the currents I1, I2 and I3 is equal to the total circuit current, I, i.e. I = I1 + I2 + I3, and • b) the source p.d., V volts, is the same across each of the resistors. • From Ohm’s law:Figure 2.3 : Parallel circuit • Source : Electrical and Electronic 8th Edition (2002)

DC Circuit • Current division • For the circuit shown in Figure 2.4, the total circuit resistance, RT is given by: Figure 2.4 : Parallel circuit • Source : Electrical and Electronic 8th Edition (2002) • 2.6, E1 is positive and E2 is negative.)

Light Circuit

DC Circuit

DC Circuit

DC Circuit

DC Circuit

DC Circuit

DC Circuit

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Kirchhoff’s laws

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Kirchhoff’s laws state: Current Law. At any junction in an electric circuit the total current flowing towards that junction is equal to the total current flowing away from the junction, Thus, referring to Figure 2.5:

or • Voltage Law. In any closed loop in a network, the algebraic sum of the voltage drops (i.e. products of current and resistance) taken around the loop is equal to the resultant e.m.f. acting in that loop. Thus, Referring to Figure 2.6: •Note that if current flows away from the positive terminal of a source, that source is considered by convention to be positive. Thus moving anticlockwise around the loop of Figure

DC Circuit • •

QUESTIONS Define the Ohm’s Law and Kirchoff’s Law. • Three resistors A = 50 Ω, B = 20 Ω & C = 25 Ω. Calculate the total resistance if the resistors connected in series and parallel. • A three resistor R1 = 2 Ω, R2 = 3 Ω and R3 = 8 Ω are connected in series. Find the voltage across each of the resistors and supply voltage if the current supply is 1.5 A. • For the network shown below I1 = 2.5 A, I2 = -1.5 A. Calculate the current I3.

DC Motor

RESISTORS IN SERIES • A simple SERIES CIRCUIT is shown in the diagram below. The current (I) at every point in a series circuit equals the current leaving the battery.

I1= I2=I3=ITotal

RESISTORS IN SERIES •

Assuming that the connecting wires offer no resistance to current flow, the potential difference between the terminals of the battery (V) equals the sum of the potential differences across the resistors, i.e., V=Vl+ V2+ V3 •The equivalent electrical resistance (R) for this combination is equal to the sum of the individual resistors, i.e., R=R1+ R2+ R3

RESISTORS IN PARALLEL In a simple PARALLEL CIRCUIT, the current leaving the battery divides at junction point A in the diagram shown below and recombines at point B. The battery current (I) equals the sum of the currents in the branches. In general I = I1 + I2 + I3

RESISTORS IN PARALLEL • If no other resistance is present, the potential difference across each resistor equals the potential difference across the terminals of the battery. • The equivalent resistance (R) of a parallel combination is always less than the smallest of the individual resistors. The formula for the equivalent resistance is as follows: • 1/R = 1/RI + 1/R2 + 1/R3 • The potential difference across each resistor in the arrangement is the same, i. e. • V = VI = V2 = V3

RESISTORS IN PARALLEL In a simple PARALLEL CIRCUIT, the current leaving the battery divides at junction point A in the diagram shown below and recombines at point B. The battery current (I) equals the sum of the currents in the branches. In general I = I1 + I2 + I3

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EMF AND TERMINAL All sources ofVOLTAGE emf have what is known as INTERNAL

RESISTANCE (r) to the flow of electric current. The internal resistance of a fresh battery is usually small but increases with use. Thus the voltage across the terminals of a battery is less than the emf of the battery. • The TERMINALVOLTAGE (V) is given by the equation V = ε - Ir, where ε represents the emf of the source of potential in volts, I the current leaving the source of emf in amperes and r the internal resistance in ohms. • The internal resistance of the source of emf is always considered to be in a series with the external resistance present in the electric circuit.

KIRCHHOFF'S RULES • KIRCHHOFF'S RULES are used in conjunction with Ohm's law in solving problems involving complex circuits: • KIRCHHOFF'S FIRST RULE or JUNCTION RULE: The sum of all currents entering any junction point equals the sum of all currents leaving the junction point. This rule is based on the law of conservation of electric charge. • KIRCHHOFF'S SECOND RULE or LOOP RULE: The algebraic sum of all the gains and losses of potential around any closed path must equal zero. This law is based on the law of conservation of energy.

SUGGESTIONS FOR USING KIRCHHOFF'S LAWS 1. Place a (+) sign next the long line of the battery symbol and a (-) sign next to the short line. Start choosing a direction for conventional current flow ( flow of positive charge ) If you choose the wrong direction for the flow of current in a particular branch, your final answer for the current in that branch will be negative. The negative answer indicates that the current actually flows in the opposite direction.

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SUGGESTIONS FOR USING KIRCHHOFF'S LAWS • 2. Assign a direction to the circuit in each independent branch of the circuit. Place a positive sign on the side of each resistor where the current enters and a negative sign on the side where the current exits, e.g.; This indicates that a drop in potential occurs as the current passes through the resistor .

SUGGESTIONS FOR USING KIRCHHOFF'S LAWS

• Notice how the directions of the currents are labeled in each branch of the circuit

SUGGESTIONS FOR USING KIRCHHOFF'S LAWS • 3. Select a JUNCTION POINT and apply the junction rule, e.g., at point A in the diagram:

The junction rule may be applied at more than one junction point. In general, apply the junction rule to enough junctions so that each branch current appears in at least one equation.

SUGGESTIONS FOR USING KIRCHHOFF'S LAWS

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4. Apply Kirchhoff’s loop rule by first taking note whether there is a gain or loss of potential at each resistor and source of emf as you trace the closed loop. Remember that the sum of the gains and losses of potential must add to zero.

SUGGESTIONS FOR USING KIRCHHOFF'S LAWS For example, for the left loop of the sample circuit above start at point B and travel clockwise around the loop. Because the direction chosen for the loop is also the direction assigned for the current, there is a gain in potential across the battery (- to +), but a loss of potential across each resistor (+ to -).

SUGGESTIONS FOR USING KIRCHHOFF'S LAWS • Following the path of the current shown in the diagram and using the loop rule, the following equation can be written:

SUGGESTIONS FOR USING KIRCHHOFF'S LAWS The direction taken around the loop is ARBITRARY. Tracing a counterclockwise path around the circuit starting at B, as shown in the above right diagram, there is gain in potential across each resistor to (- to +) and a drop in potential across the battery (+ to -). The loop equation would then be:

SUGGESTIONS FOR USING KIRCHHOFF'S LAWS • Multiplying both sides of the above equation by - 1 and algebraically rearranging, it can be shown that the two equations are equivalent. Be sure to apply the loop rule to enough closed loops so that each branch current appears in at least one loop equation. Solve for each branch current using standard algebraic methods. “Solve simultaneous equations”

CAPACITORS IN SERIES AND PARALLEL •

A circuit with CAPACITORS IN PARALLEL is shown in the diagram below. According to Kirchhoff ‘s loop rule, the potential difference (V) of the source of emf:

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V = VI = V2 = V3

CAPACITORS IN PARALLEL • • • • •

The total charge stored on the capacitor plates (Q) equals the amount of charge which left the source of: Q = Ql + Q2 + Q3 ( Charge is additive) and since Q = CV then CV = CV1 + CV2 + CV3 C= C1+ C2 +C3 (Capacitance is additive)

CAPACITORS IN SERIES •

For CAPACITORS IN SERIES, the amount of charge (Q) that leaves the source of emf equals the amount of charge that forms on each capacitor:

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Q = Ql = Q2 = Q3

CAPACITORS IN SERIES •

From Kirchhoffs loop rule, the potential difference across the source of emf (V) equals the sum of the potential differences across the individual capacitors:

Circuits containing resistors and capacitors An RC CIRCUIT consists of a resistor and a capacitor connected in series to a de power source.When switch 1 (S1), shown in the diagram below, is closed, the current will begin to flow from the source of emf and charge will begin to accumulate on the capacitor. Using Kirchhoff s loop rule it can be shown that