This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full
Document
This homework help was uploaded on 04/19/2008 for the course PHYS 2074 taught by Professor Stewart during the Spring '08 term at Arkansas.
 Spring '08
 Stewart
 Physics, Work

Click to edit the document details