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PHYS 012 1st Edition Lecture 10 Outline of Last Lecture II. 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.III. 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 ΩThese 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.c. 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 ΩIV. 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.V. 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 + … + 1/Cn.Outline of Current Lecture VI. Kirchhoff’s Rulesa. Junction rule: the total current going into a junction equals the total current out of the junctioni. ΣIin = ΣIoutii. Derived from the law of conservation of chargeb. Loop rule: the sum of changes in potential around a closed circuit loop is zeroi. ΣΔVclosed loop = 0c. Problem: Find the values for current through the closed circuit below given: ε1 = 10V, ε2 = 14V, R1 = 6Ω, R2 = 4Ω, R3 = 2Ω.i. First, draw currents through this circuit. Direction does not matter at this point in the problem. When you get values for the current, if a value is negative then current flows in the opposite direction as predicted. The magnitude will still be the same.1. Since ΣIin = ΣIout, for junction A, I1 + I2 = I3.ii. Second, mark each side of every element in the circuit with higher (+) or lower (-) potential. Remember: current flows from high potential to low potential.iii. Third, add up the potentials around each part of the circuit (top and bottom). The potentials should add up to zero (loop rule). The magnitudeof the change in potential will be negative if you’re going from lower (-) tohigher (+) potential.1. Following top loop clockwise: +I1R1 – ε1 – I2R2 – ε2 = 02. Following bottom loop clockwise: -I3R3 + ε1 – I1R1 = 0iv. Finally, solve for I1, I2, and I3.1. Substituting the values for R1, R2, ε1, and ε2: +6I1 – 10 – 4I2 – 14 = 6I1 - 4I2 – 24 = 02. Substituting the values for R1, R3, and ε1: -2I1 – 2I2 + 10 – 6I1 = -8I1 –2I2 + 10 = (- 1/2)(16I1 + 4I2 - 20) = 03. Adding the two equations together: 22I1 – 44 = 04. Solve for I1: I1 = 44/22 = 2.0 A5. Solve for I2: 6I1 - 4I2 – 24 = 0; 6(2) - 4I2 – 24 = 0; I2 = 12/(-4) = -3.0 A6. Solve for I3: I1 + I2 = I3 ; I3 = (2.0) + (-3.0) = -1.0 AVII. RC Circuits:a. Chargingi. ΔV = Q/Cii. When battery is first connected to the circuit, voltage across the capacitorincreases from zero while voltage across the resistor decreases until the voltage across the resistor is zero and current stops.iii. Time constant (τ) = RC1. When t = τ, I is reduced by a factor of 1/e.2. After a few time constants, I is assumed to be zero, even though it never actually reaches zero.b. Dischargingi. When the battery is removed from the circuit, electric potential energy is stored in the capacitor, which then produces a current in the opposite direction.ii. Electric potential energy is gradually lost through the resistors in the formof thermal energy.c. Problem: Find I at t = 0 (right after closing switch) and as t approaches ∞ (after switch has been closed for a long time) in the circuit below.i. When t = 0, Iε = ε/R1.ii. When t ∞, Iε =

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