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Lecture 20 Lc & Rlc Circuits & Ac Circuits I

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Lecture 20 Lc & Rlc Circuits & Ac Circuits I LECTURE 20 LC & RLC CIRCUITS & AC CIRCUITS I Lecture 20 2 ¨ Reading chapter 29-1 and 29-3 to 29-4 ¤ LC circuits ¤ RLC circuits (without AC generator) ¤ Transformers ¤ AC circuit with resistor ¤ RMS LC circuits 3 ¨ Consider an LC circuit with initial charge Q0 stored in the upper capacitor plate and with the switch open. At t = 0, the switch is closed. )* , )-, , ¨ � + � = 0 ⇒ � + = 0 ⇒ � + = 0 " $ )+ $ )+- $ 1 Q t Q cos t, where ( ) = 0 ω ω = LC dQ I t Q sin t ( ) = = −ω 0 ω dt ¨ Q lags I by 90°. ¨ Total energy remains constant. Energy in LC circuits 4 ¨ The sum of the electric and magnetic energies stored in a LC circuit at any given time is the same as the initial electric energy stored in the capacitor. ¨ Energy does not dissipate through capacitors or inductors. 1 Q2 1 1 Q2 U + U = + LI 2 = 0 e m 2 C 2 2 C +Q0 -Q0 -Q0 +Q0 Example: 1 5 ¨ You have a 10 mH inductor. What capacitor should you use with it in series to make an oscillator with a frequency of 920 kHz? (This frequency is near the center of the AM radio band.) Demo: 1 6 ¨ Tesla coil ¤ Produces 0.5 million volts at 350 kHz ¤ Demonstration of resonance Spark gap L L’ Lp L Cs 110Vac Cp s ZAP! 110V to 15kV ac ac 1 1 transformer ω0 = = L C L C p p s s RLC circuits 7 ¨ Consider an RLC circuit with an initially charged capacitor and with the switch open. After the switch is closed, if R is small, Q and I oscillate with the natural frequency, ω0, but the oscillation is damped. 1 ω0 = LC ¨ The initial energy stored in the capacitor dissipates through the resistor. The higher the resistance of the resistor, the more damping the circuit has. Demo: 2 8 ¨ LC oscillation ¨ Damped RLC oscillation Alternating current 9 ¨ When you plug a lamp into a wall socket, the voltage and current supplied to the light bulb vary sinusoidally. ¨ The voltage oscillates with a fixed frequency of 60 cycles per second in the US. ¨ The current therefore also periodically reverses direction, so the wall socket provides an alternating current (ac). Quiz: 1 10 AC generator 11 ¨ An alternating current (ac) generator produces an emf given by E = Epeak cosωt where Epeak is the largest value attained by the voltage during a cycle, ω is the angular frequency, and t is time. ¨ In circuit diagrams, an ac generator is represented by the symbol: Resistors in AC circuits 12 ¨ The potential drop across the resistor, VR, and current through the circuit, I, are given by VR = VR peak cosωt V VR R peak I = = cosωt ≡ Ipeak cosωt R R ¨ VR and I are in phase. rms current and voltage 13 ¨ The voltage and current in an ac circuit both average to zero, making the average useless in describing their behavior. ¨ We use instead the root mean square (rms). 5 2 ¨ In general, the rms value of x is given by �/01 = � 34 ¨ The rms values of sinusoidally varying current and voltage are given by: I max Vmax Irms = V = 2 rms 2 ¤ 120 volts is the rms value of household ac. ¨ The rms or max current is related to the rms or max voltage by Vrms Vmax Irms = Imax = R R Power delivered to the resistor 14 ¨ The power, P, and average power, Pav, delivered to the resistor are given by 2 2 2 P = I R = Ipeak Rcos ωt 1 P = I 2 R = I 2 R av ( )av peak 2 P = I 2 R = E I av rms rms rms Quiz: 2 15 Example: 2 16 ¨ A resistor with a resistance of R = 3.33 kΩ is connected to a generator with a peak voltage of Vpeak = 141 V. a) Find the average power delivered to this circuit. b) Find the maximum power delivered to this circuit. Transformers 17 ¨ A transformer is a device used to increase or decrease the voltage in a circuit without a loss of power. ¨ The voltages across the primary and secondary coils are related by the numbers of turns of the coils. ¨ Since the power in both circuits must be the same, if the voltage is lower, the current must be higher. � � � 6 = 6 = 2 /01 �2 �2 �6 /01 V2 Quiz: 3 through 5 18 Demo: 3 19 ¨ Transformers ¤ Turn ratios of 1:1, 1:2, and 2:1. Example: 3 20 ¨ The primary coil of a step-down transformer has N1 = 250 turns and is connected to an outlet of V1 = 120 V. The secondary coil is to supply I2 rms = 20 A at V2 = 9.0 V. a) Find the rms current in the primary coil. b) Find the number of turns in the secondary coil. Demo: 4 21 ¨ Transformers and power lines ¤ To minimize the P = I2R heat loss in electric transmission lines, we use a high voltage and a low current. ¤ For safety, power is delivered to consumers at lower voltages and therefore with higher currents. ¤ Because of this ease of stepping the voltage up or down with transformers, AC rather than DC is in common use. Electrical fuse and circuit breaker 22 ¨ Electrical fires can be started by improper or damaged wiring because of the heat caused by a too-large current or resistance. ¨ A fuse is designed to be the hottest point in the circuit – if the current is too high, the fuse melts. ¨ A circuit breaker is similar, except that it is a bimetallic strip that bends enough to break the connection when it becomes too hot. When it cools, it can be reset. Polarized and grounded plugs 23 ¨ ~0.1 A of current through a body could be fatal. ¨ With a polarized plug, the case of an appliance is always connected to the low-potential side, essentially at ground. The high-potential extends only from the wall outlet to the switch when the appliance is turned off. Polarized plug ¨ With a three-prong grounded plug, the round prong is connected to the ground and the case of an appliance. If a high-potential wire accidentally touches the case, the current flows through the third prong. Grounded plug GFCI outlets 24 ¨ The ground fault circuit interrupter (GFCI) protects against electric shock. ¤ Wire 1: from the outlet to the appliance ¤ Wire 2: from the appliance to the outlet ¨ If there is no current leakage, the net current though the iron ring is zero. ¨ If there is a leakage, the current in wire 2 is less, and sensing coil detects the change in magnetic flux. .
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