Ch33 - Physics 4B Lecture Notes Chapter 33 -...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 4B Lecture Notes 33-1 Chapter 33 - Electromagnetic Oscillations and Alternating Current Problem Set #12 - due: Ch 33 - 3, 10, 17, 18, 31, 34, 39, 42, 56, 59, 61, 77, 80, 84, 85, 87 Lecture Outline 1. The LC Circuit 2. The LRC Circuit 3. Basic AC Circuits 4. Frequency Filtering Circuits 5. The RLC Tuning Circuit 6. Power in AC Circuits 7. Transformers We will use the loop and junction theorems to analyze the behavior of circuits that exhibit oscillatory behavior. We will start by allowing the circuits to oscillate at their “natural” frequency, then we will examine the effect of forcing them to oscillate at other frequencies. 1. The LC Circuit In the circuit at the left, the capacitor is charged when the switch is connected from a to b. Then the switch is connected from b to c. The capacitor begins to lose its charge as it attempts to get current through the inductor. The inductor begins to build up a magnetic field. This field peaks as the charge on the capacitor goes to zero. Then the energy in the inductor begins to create current to charge the capacitor back up. All of this is explained by the loop theorem which requires that the potential difference across the capacitor always equal the potential difference across the inductor, V c = V L Q C = - L dI dt Q C = - L d 2 Q dt 2 d 2 Q dt 2 = - 1 LC Q This differential equation is the SHM equation, d 2 x dt 2 = - k m x = -ϖ 2 x x = Acos( ϖ t ) . By analogy, Q = Q m cos( ϖ t ) The LC Circuit where ϖ = 1 LC Example 1: Show that the above expression for the charge on the capacitor satisfies the loop theorem. Q = Q m cos( ϖ t ) dQ dt = - Q m ϖ sin( ϖ t ) d 2 Q dt 2 = - Q m ϖ 2 cos( ϖ t ) = -ϖ 2 Q Substituting into the loop theorem, 2 / Q = - 1 LC / Q ⇒ ϖ = 1 LC . L ε C a b c
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Physics 4B Lecture Notes 33-2 Example 2: For an LC circuit find (a)the peak current in terms of the peak charge on the capacitor and (b)the fraction of the period between the time the capacitor's charge peaks and the current peaks. (a)From the definition of current, I dQ dt = - Q m ϖ sin( ϖ t ) I m = Q m ϖ . (b)First let's define Q=Q m when t=0. This establishes the value of δ , Q = Q m cos( ϖ t ) Q m = Q m cos( δ ) cos( δ ) = 1 ⇒ δ = 0 Note that I = I m when sin( ϖ t ) = 1 ⇒ ϖ t = π 2 ⇒ ϖ t = π 2 t = π 2 ϖ . Recall, ϖ = 2 π T t = π 2 2 π T t = 1 4 T . 2. The LRC Circuit The capacitor is charged by connecting a to b. Then the switch is moved from b to c. The oscillations of the LC circuit are damped out by the heat created in the resistor. As usual, the loop theorem gives the details, V L + V R + V C = 0 L dI dt + IR + Q C = 0 L d 2 Q dt 2 + R dQ dt + Q C = 0 If we assume that R is fairly small then we would expect damped oscillations of the form, Q = Q m e t cos ϖ d t where Q m is the initial charge on the capacitor, α is the damping constant, and ϖ d is the frequency of oscillation. A graph of
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 12

Ch33 - Physics 4B Lecture Notes Chapter 33 -...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online