Physics 012 1st Edition Lecture 9 Outline of Last Lecture I. Electric current: net flow of charge per unit time through some regiona. By convention, current flows in the direction that positive charge would move. Electrons (the charge carriers for metals) will flow in the opposite direction of electric current.b. Measured in ampheres, amps (A)i. 1 A = 1 C/sII. Ohm’s Law: relationship between potential difference and currenta. Drift speed (vd): net speed an electron movesi. Path of motion is not actually linear due to collisions with other particles, which causes a certain amount of resistance as electrons move through a material.b. ΔV = IRi. R = resistance = (ρL)/A for a conductive wire1. ρ = resistivity (larger with more resistance, like in non-metals)2. L = length of conductive wire3. A = cross-sectional area of wire (πr2)c. Only applicable to ohmic materialsd. R = ΔV/I in ohms (Ω)i. 1 Ω = 1 V/Ae. ρ in Ω*mIII. Electrical power: rate of energy consumed by resistora. P = IΔV ( = I2R = V2/R )i. V = IR ; I = V/RThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.b. Electric circuits: i. Basic circuit- wire in picture (connecting battery and resistor) is an ideal conducting wire (no resistance)ii. In series- In general, Req. = R1 + R2 + … + RnOutline of Current Lecture IV. Parallel Resistorsa. ΔV = ΔV1 = ΔV2b. I = I1 = I2c. V = IRd. ΔV/Req = ΔV1/R1 + ΔV2/R2e. 1/Req = 1/R1 + 1/R2f. In general, for parallel resistors, 1/Req = 1/R1 + 1/R2 + … + 1/Rn.V. Problem: For the circuit below, find a single equivalent resistance.a. The 4 Ω and 6 Ω resistors are in series so Req,4,6 = 4 Ω + 6 Ω = 10 Ωb. The 9 Ω, 8 Ω, and “10 Ω” resistors are parallel, so Req, 9, 8, 10 = 1/8 Ω + 1/9 Ω + 1/10 Ω = 2.98 Ωc. The 3 Ω and “2.98 Ω” resistors are in series, so Req, 3, 2.98 = 3 Ω + 2.98 Ω = 5.98 Ωd. The 20 Ω and the “5.98 Ω” resistors are parallel, so Req = 20 Ω + 5.98 Ω = 25.98 ΩVI. Parallel Capacitorsa. ΔV = ΔV1 + ΔV2b. q = q1 = q2c. q = CVd. ΔVCeq = ΔV1C1 + ΔV2C2e. Ceq = C1 + C2f. In general, for parallel capacitors, Ceq = C1 + C2 + … + Cn.VII. Capacitors in Seriesa. q = q1 = q2b. ΔV = ΔV1 + ΔV2c. V = q/Cd. q/Ceq = q/C1 + q/C2e. 1/Ceq = 1/C1 + 1/C2f. In general, for capacitors in series, 1/Ceq = 1/C1 + 1/C2 + … +
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