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Unformatted text preview: Solution for Homework 13 RC and Review for Exam 2 (Required but not Collected) Solution to Homework Problem 13.1(Time Dependence of Decaying Exponential) Problem: Given the plot below of the voltage across a resistor through which a capacitor is discharging: (a)What is the functional form of the time dependence? (b)What is the time constant? (c)How long does it take for V R to decay to 10 % of its value at t = 0 ? 5 10 15 20 25 30 t (sec) 1 2 3 4 5 6 7 8 9 10 V R (t) (Volts) V Solution to Part(a) The time dependence is a decaying exponential, V R ( t ) = V e t , where V = 10 V and = RC is the time constant. Grading Key: Part (a) 2 Points Solution to Part(b) The time constant is the time where the voltage reaches V R ( ) = V e 1 = 0 . 37 V . I read this time as = 10s from the graph. 1 Grading Key: Part (b) 2 Points Solution to Part(c) 10% of V is 1 V . The plot reaches 1 V at t = 22 . 5s . Grading Key: Part (c) 2 Points Total Points for Problem: 6 Points Solution to Homework Problem 13.2(Two Loop Kirchhoff Law Problem) Problem: For the circuit at the right with current directions as drawn, V 1 = 12 V , V 2 = 6 V , and all resistors are 100 . Calculate the currents, I 1 , I 2 , and I 3 using the loops to the right. a b c d e f Loop 1 Loop 2 V 2 V 1 I 1 I 3 I 2 Solution a b c d e f V 2 V 1 I 1 I 3 I 2 R s Loop 1 Loop 2 Definitions R = 100 Resistance of Each Resistor R s Resistance of Series Combination I i Current V i Potential Difference of Battery i 2 Strategy: Draw two independent loops. Write Loop equations for each loop and one junction equation. Solve the simultaneous system for the currents. (a) Reduce Series Combination: The two resistors through which I 2 passes are in series, replace them with their equivalent, R s = R + R = 2 R = 200 , and redraw the circuit as shown above. (b) Write Junction Equation: Since charge is conserved, the charge flowing into a junction equals the charge leaving the junction; therefore, I 1 + I 3 = I 2 . (c) Write Loop Equations: The sum of the potential drops around any closed path is zero, so we can write the following loop equations. V 1 I 1 R I 2 R s = 0 Loop 1 I 3 R V 2 + I 2 R s = 0 Loop 2 (d) Eliminate I 3 using Junction Equation: Substitute the junction equation into loop 2 to eliminate I 3 = I 2 I 1 , giving ( I 2 I 1 ) R V 2 + I 2 R s = 0 Eqn 3 . Use the fact that R s = 2 R and collect terms in the current. I 1 R V 2 + 3 RI 2 = 0 Eqn 4 (e) Compute I 2 : Subtract Eqn 4 from Loop 1 equation to give, V 1 I 2 R s + V 2 3 RI 2 = V 1 + V 2 5 I 2 R = 0 or I 2 = V 1 + V 2 5 R = 12 V + 6 V 5(100) = 18 V 500 = 0 . 036 A....
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This note was uploaded on 03/29/2010 for the course PHYS 2469 taught by Professor Stewat during the Spring '10 term at University of Arkansas Community College at Batesville.
 Spring '10
 Stewat
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