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Chapter 20 Electric Circuits

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A battery consists of chemicals, called electrolytes, sandwiched in between 2 electrodes, or terminals, made of different metals. Chemical reactions do work to separate charge, which creates a potential difference between positive and negative terminals A physical separator keeps the charge from going back through the battery. The Battery

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ε and ∆V For an ideal battery potential difference between the positive and negative terminals, ∆V, equals the chemical work done per unit charge. ∆V = W ch /q = ε ε is the emf of the battery Due to the internal resistance of a real battery, ∆V is often slightly less than the emf.

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How do we know there is a current?

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20.1 Electromotive Force and Current The electric current is the amount of charge per unit time that passes through a surface that is perpendicular to the motion of the charges. One coulomb per second equals one ampere (A).

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20.1 Electromotive Force and Current If the charges move around the circuit in the same direction at all times, the current is said to be direct current (dc). If the charges move first one way and then the opposite way, the current is said to be alternating current (ac).

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20.1 Electromotive Force and Current Example 1 A Pocket Calculator The current in a 3.0 V battery of a pocket calculator is 0.17 mA. In one hour of operation, (a) how much charge flows in the circuit and (b) how much energy does the battery deliver to the calculator circuit?

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20.1 Electromotive Force and Current Example 1 A Pocket Calculator The current in a 3.0 V battery of a pocket calculator is 0.17 mA. In one hour of operation, (a) how much charge flows in the circuit and (b) how much energy does the battery deliver to the calculator circuit? (a) (b)

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20.1 Electromotive Force and Current Conventional current is the hypothetical flow of positive charges (electron holes) that would have the same effect in the circuit as the movement of negative charges thatactually does occur.

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20.2 Ohm’s Law The resistance (R) is defined as the ratio of the voltage V applied across a piece of material to the current I through the material.

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20.2 Ohm’s Law OHM’S LAW The ratio V/I is a constant, where V is the voltage applied across a piece of mateiral and I is the current through the material: SI Unit of Resistance: volt/ampere (V/A) = ohm (Ω)

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20.2 Ohm’s Law To the extent that a wire or an electrical device offers resistance to electrical flow, it is called a resistor.

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20.2 Ohm’s Law Example 2 A Flashlight The filament in a light bulb is a resistor in the form of a thin piece of wire. The wire becomes hot enough to emit light because of the current in it. The flashlight uses two 1.5-V batteries to provide a current of 0.40 A in the filament. Determine the resistance of the glowing filament.

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Ratio Reasoning using Ohm’s Law In circuit #1, the battery has twice as much potential difference as the battery in circuit #2. However, the current in circuit # 1 is only ½ the current in circuit #2. Therefore R 1 is: A. twice D. four times B. one half E. 1/4 C. equal to the resistance R 2

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Ratio Reasoning using Ohm’s Law In circuit #1, the battery has twice as much potential difference as the battery in circuit #2. However, the current in circuit # 1 is only ½ the current in circuit #2. Therefore R 1 is: A. twice D. four times B. one half E. 1/4 C. equal to the resistance R 2 Using ratio reasoning (R 1 /R 2 ), the answer is D

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20.4 Electric Power ELECTRIC POWER Charges acquire EPE due to the potential difference across the battery. In the circuit, this energy is transformed into KE, as the charges move. The rate at which energy is delivered is Power. P = Energy/∆t When there is current in a circuit as a result of a voltage, the electric power delivered to the circuit is: SI Unit of Power: watt (W), where one watt equals one joule per second Many electrical devices are essentially resistors, so we use 2 formulas for power based on resistance. From Ohm’s Law:

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20.4 Electric Power Example 5 The Power and Energy Used in a Flashlight In the flashlight, the current is 0.40A and the voltage is 3.0 V. Find (a) the power delivered to the bulb and (b) the energy dissipated in the bulb in 5.5 minutes of operation. In this case, the resistance of the bulb is not given, so this determines the formula for power.

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20.4 Electric Power (a) (b)

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Power rating of a light bulb or household applicance Commercial and residential electrical systems are set up so that each individual appliance operates at a potential difference of 120 V. Power Rating or Wattage is the power that the appliance will dissipate at a potential difference of 120 V. Power consumption will differ if operated at any other potential difference (i.e. 220, such as is standard in Europe and many other countries).

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Resistors in Series The current, I, is the same through each of the resistor. The potential difference through each resistor is : ∆V R = (-)IR The equivalent resistor is made by adding up all resistors in series in the circuit. ∆V bat = (-) ∆V circuit = (-)IR s Why the negative sign?

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20.6 Series Wiring Current is a vector. Positive direction is from positive to negative terminal. ΔV bat is positive; positive terminal has more potential than negative. ΔV R is negative; it is higher at the initial value than at the final By understanding that the potential difference across any circuit resistor will be a voltage DROP, while the potential difference back through the battery (charge escalator) will be a GAIN, one can replace the ΔV with “V”, and dispense with the negative sign in Ohm’s Law. V 1 (voltage drop across the resistor1) = IR 1 V 2 (voltage drop across the resistor2) = IR 2 V bat = V cir = IR eq

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20.6 Series Wiring Example 8 Resistors in a Series Circuit A 6.00 Ω resistor and a 3.00 Ω resistor are connected in series with a 12.0 V battery. Assuming the battery contributes no resistance to the circuit, find (a) the current, (b) the power dissipated in each resistor, and (c) the total power delivered to the resistors by the battery.

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20.6 Series Wiring (a) (b) (c)

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Power in series A 60 W light bulb and a 100 W bulb are placed one after the other in a circuit. The battery’s emf is 120V. Which one glows more brightly and why? a.100 W because more current flows through it b. 100 W because of the higher potential difference across the bulb c.60 W because more current flows through it d.60 W because of the higher potential difference across the bulb

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Power in series A 60 W light bulb and a 100 W bulb are placed one after the other in a circuit. The battery’s emf is 120V. Which one glows more brightly and why? a.100 W because more current flows through it b. 100 W because of the higher potential difference across the bulb c.60 W because more current flows through it d.60 W because of the higher potential difference across the bulb

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20.7 Parallel Wiring Parallel wiring means that the devices are connected in such a way that the same voltage is applied across each device. When two resistors are connected in parallel, each receives current from the battery as if the other was not present. Therefore the two resistors connected in parallel draw more current than does either resistor alone.

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20.7 Parallel Wiring

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The two parallel pipe sections are equivalent to a single pipe of the same length and same total cross sectional area.

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20.7 Parallel Wiring parallel resistors note that for 2 parallel resistors this equation is equal to : R p = (R 1 R 2 ) / (R 1 + R 2 )

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20.7 Parallel Wiring Example 10 Main and Remote Stereo Speakers Most receivers allow the user to connect to “remote” speakers in addition to the main speakers. At the instant represented in the picture, the voltage across the speakers is 6.00 V. Determine (a) the equivalent resistance of the two speakers, (b) the total current supplied by the receiver, (c) the current in each speaker, and (d) the power dissipated in each speaker.

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20.7 Parallel Wiring (a) (b)

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20.7 Parallel Wiring (c) (d)

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Power in parallel A 60 W light bulb and a 100 W bulb are placed in parallel in a circuit. The battery’s emf is 120V. Which one glows more brightly and why? a.100 W because more current flows through it b. 100 W because of the higher potential difference across the bulb c.60 W because more current flows through it d.60 W because of the higher potential difference across the bulb 60W 100W

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Power in parallel A 60 W light bulb and a 100 W bulb are placed in parallel in a circuit. The battery’s emf is 120V. Which one glows more brightly and why? a.100 W because more current flows through it b. 100 W because of the higher potential difference across the bulb c.60 W because more current flows through it d.60 W because of the higher potential difference across the bulb

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Circuits Wired Partially in Series and Partially in Parallel Example 12 Find a) the total current supplied by the the battery and b) the voltage drop between points A and B.

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Circuits Wired Partially in Series and Partially in Parallel Example 12 a)I tot = V/R p = 24V/240Ω =.10A b)now go back to the 1st circuit in part c to calculate the voltage drop across A-B: V AB = IR AB = (.10A)(130 Ω) = 13V

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Rank the bulbs Rank the brightness of bulbs. “Out” is the least bright. Parentheses show ties a.A, B, C b.A, (B, C) c.(A, B), C d.(ABC) http://bcs.wiley.com/he- bcs/Books?action=resource&bcsId=4678&ite mId=0470223553&resourceId=15300

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Rank the bulbs Rank the brightness of bulbs. “Out” is the least bright. Parentheses show ties a.A, B, C b.A, (B, C) c.(A, B), C d.(ABC) http://bcs.wiley.com/he- bcs/Books?action=resource&bcsId=4678&ite mId=0470223553&resourceId=15300

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What happens to A? With switch closed, A is____________: a.Brighter than b.Less bright than c.Equal to what it was when switch was open. Use Ohm’s Law http://bcs.wiley.com/he- bcs/Books?action=resource&bcsId=4678&ite mId=0470223553&resourceId=15300 closed

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What happens to A? With switch closed, A is____________: a.Brighter than b.Less bright than c.Equal to what it was when switch was open. Use Ohm’s Law http://bcs.wiley.com/he- bcs/Books?action=resource&bcsId=4678&ite mId=0470223553&resourceId=15300 closed

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Your Understanding What is the ratio of the power supplied by the battery in parallel circuit A to the power supplied by the battery in series circuit B? a.¼ c. 2e. 1 b. 4 d. 2 A. B.

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Your Understanding What is the ratio of the power supplied by the battery in parallel circuit A to the power supplied by the battery in series circuit B? a.¼ c. ½ e. 1 b.4 d. 2 A. B.

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20.9 Internal Resistance Batteries and generators add some resistance to a circuit. This resistance is called internal resistance. The actual voltage between the terminals of a battery is known as the terminal voltage. Recall the ideal voltage is the EMF

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20.9 Internal Resistance Example 12 The Terminal Voltage of a Battery The car battery has an emf of 12.0 V and an internal resistance of 0.0100 Ω. What is the terminal voltage when the current drawn from the battery is (a) 10.0 A and (b) 100.0 A? (a) (b)

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20.11 The Measurement of Current and Voltage A dc galvanometer. The coil of wire and pointer rotate when there is a current in the wire.

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20.11 The Measurement of Current and Voltage An ammeter must be inserted into a circuit so that the current passes directly through it.

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20.11 The Measurement of Current and Voltage If a galvanometer with a full-scale limit of 0.100 mA is to be used to measure the current of 60.0 mA, a shunt resistance must be used so that the excess current of 59.9 mA can detour around the galvanometer coil.

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20.11 The Measurement of Current and Voltage To measure the voltage between two points in a circuit, a voltmeter is connected between the points.

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Series Resistors What is the value of R (3 rd resistor in circuit)? a. 50 Ω b. 25Ω c. 10 Ω d. Other

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Series Resistors What is the value of R (3 rd resistor in circuit)? a. 50 Ω b. 25Ω c. 10 Ω d. Other R s = R 1 + R 2 + R 3 = ΔV/I.

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