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Transcript
Electromotive Force The maximum potential difference between the terminals on a source of electric energy is called the electromotive force (EMF) and has units of volts Symbol for EMF in a circuit diagram Electric Current The electric current is the amount of charge that passes through a cross-section of a wire per second Ξπ πΌ= Ξπ‘ Unit: C/s = A (ampere) Conventional Current Conventional current is the hypothetical flow of positive charges through the circuit that has the same effect as the flow of negative charges that actually occurs. Direct current is a current that only flows in one direction through the circuit. Alternating current is a current that varies in magnitude and reverses direction periodically. Resistance Resistance is the ratio of the voltage V applied across a material to the current running through the material π = π πΌ π’πππ‘: π π΄ = Ξ© (ohm) This relationship is called Ohmβs Law. The resistance is a constant. Ohmβs Law is usually written as V=IR Resistor resistor Any device that offers resistance to the flow of charges is called a resistor emf Circuit diagram Calculating Resistance For a wire that has a length, L, and a cross-sectional area, A, the resistance can be calculated from πΏ π =Ο π΄ Where Ο is called the resistivity (unit=Ξ©m) and is a property of the material the wire is made from. Conductors have a very small resistivity while insulators have a very large resistivity. Materials that are in between a conductor and insulator are called semiconductors. Table of Resistivities Temperature Dependence of Resistivity and Resistance ο² ο½ ο²o ο1 ο« ο‘ ο¨T ο To ο©ο temperature coefficient of resistivity R ο½ Ro ο1 ο« ο‘ ο¨T ο To ο©ο Electric Power In a time of Ξt, a charge of Ξq will pass through the battery πΈπππππ¦ Ξππ π= = = πΌπ π‘πππ Ξπ‘ The power dissipated by the resistor is 2 π π = πΌπ = πΌ2 π = π π’πππ‘: π (π€ππ‘π‘) Practice Questions Question 1: A steady current of 0.75A flows through a wire. How much charge passes through the wire in 1.5 minutes? Question 2: How many electrons flow through a light bulb each second if the current through the light bulb is 0.75A? Question 3: A metal rod is 2 m long and 8 mm in diameter. Compute its resistance if the resistivity of the metal is 1.76 X 10-8 β¦ m. Question 4: An ammeter-voltmeter method is used to measure an unknown resistance R. The ammeter reads 0.3 A, and the voltmeter reads 1.5 V. Compute the value of R (neglecting the internal resistance of the ammeter and voltmeter). Question 5: An aluminum wire has a diameter of 2.59 mm. What is the length of wire required to give a resistance of 1 β¦. Ο for aluminum is 2.8 X 10-8 β¦ m. Question 6: The resistance of a coil of copper wire is 3.5 β¦ at 0oC. What is the resistance at 50oC. For copper, Ξ± = 4.3 X 10-3 oC-1. Alternating Current π = π0 sin(2Οππ‘) V0 is the peak voltage and f is the frequency of the oscillations. If the circuit only contains resistance the current is given by π π0 πΌ = = sin 2Οππ‘ = πΌ0 sin(2Οππ‘) π π Power in Alternating Current π = πΌπ π = (πΌ0 sin 2Οππ‘ )(π0 sin(2Οππ‘)) π = πΌ0 π0 π ππ2 (2Οππ‘) I oVo ο¦ I o οΆο¦ Vo οΆ Pο½ ο½ο§ ο·ο§ ο· ο½ I rmsVrms 2 ο¨ 2 οΈο¨ 2 οΈ Ohmβs Law for Alternating Current Vrms = Irms R P = Irms Vrms 2 P = Irms R 2 ππππ π= π