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Unformatted text preview: Physics 1B: Electricity & Magnetism Summer Session I, 2010 Quiz #4, July 22 This is version A ! Useful coefficients: Mass of proton, m p = 1 . 67 × 10 − 27 kg; Mass of electron, m e = 9 . 11 × 10 − 31 kg Unit of elementary charge e = 1 . 6 × 10 − 19 C Coulomb’s constant, k e = 8 . 99 × 10 9 N m 2 C − 2 Permittivity of free space ǫ = 1 4 πk e = 8 . 85 × 10 − 12 C 2 N − 1 m − 2 1 eV = 1.6 × 10 − 19 J g = 9 . 8 m s − 2 1 T = 10 4 G 1. Three resistors, each with a resistance of 4Ω, are connected in parallel to a 12V battery; what’s the current through any one resistor? a. 0.33 A b. 1.0 A c. 3.0 A d. 12.0 A e. 9.0 A Solution: Calculate the equivalent resistance: 1 R eq = 1 4Ω + 1 4Ω + 1 4Ω = 3 4Ω So R eq = 4 3 Ω Let’s label the current flowing through each resistor as I 1 , I 2 and I 3 , so that the current flowing through the battery = I tot = I 1 + I 2 + I 3 . I tot = EMF / R eq = 12V/1.333Ω = 9.0 A. Since each resistor has equal resistance, the currents flowing through each of them must be equal: I 1 = I 2 = I 3 = I tot /3 = 3.0 A 2. A light bulb with a resistance of 220Ω, a battery with EMF=10 V, and a capacitor with C=4 milliFarads are all connected in series to form a charging RC circuit. What is the value of the voltage drop across the capacitor at a time 2.64 seconds after the switch is closed? a. 3.68 V b. 2.64 V c. 36.8 V d. 6.32 V e. 9.50 V Solution: First, we calculate the time constant τ = RC = (220Ω)(4 × 10 − 4 F) = 0.88 sec When a capacitor is charging, the voltage drop across the plates ΔV C goes from 0 to its maximum value, ǫ = EMF, following ΔV C (t) = ǫ (1 e − t/τ ) 1 At t=2.64 seconds (note that this is 3 time constants), we have ΔV C (t) = 10V(1 e − 2 . 64 s/ . 88 s ) = 10V(1 e − 3 ) = 10V(1 – 0.050) = 10V(0.95) = 9.50 V, i.e., the capacitor has reached 95%) = 10V(1 – 0....
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This note was uploaded on 09/13/2010 for the course PHYS PHYS 1B taught by Professor Berman during the Summer '09 term at UCSD.
 Summer '09
 Berman
 Charge, Magnetism, Mass

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