_012 - Solution for Homework 13 RC and Review for Exam 2...

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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 Δ V R to decay to 10 % of its value at t = 0 ? 0 5 10 15 20 25 30 t (sec) 0 1 2 3 4 5 6 7 8 9 10 V R (t) (Volts) V 0 Solution to Part(a) The time dependence is a decaying exponential, Δ V R ( t ) = Δ V 0 e - t τ , where Δ V 0 = 10 V and τ = 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 Δ V R ( τ ) = Δ V 0 e - 1 = 0 . 37Δ V 0 . I read this time as τ = 10s from the graph. 1
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Grading Key: Part (b) 2 Points Solution to Part(c) 10% of Δ V 0 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 KirchhoF Law Problem) Problem: For the circuit at the right with current directions as drawn, Δ V 1 = 12 V , Δ V 2 = 6 V , and all resistors are 100Ω . 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 Δ V 2 Δ V 1 I 1 I 3 I 2 Solution a b c d e f Δ V 2 Δ V 1 I 1 I 3 I 2 R s Loop 1 Loop 2 De±nitions R = 100Ω Resistance of Each Resistor R s Resistance of Series Combination I i Current Δ V i Potential Di²erence of Battery i 2
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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 = 200Ω , and redraw the circuit as shown above. (b) Write Junction Equation: Since charge is conserved, the charge Fowing 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. Δ V 1 - I 1 R - I 2 R s = 0 Loop 1 I 3 R - Δ 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 - Δ 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 - Δ V 2 + 3 RI 2 = 0 Eqn 4 (e) Compute I 2 : Subtract Eqn 4 from Loop 1 equation to give, Δ V 1 - I 2 R s + Δ V 2 - 3 RI 2 = Δ V 1 + Δ V 2 - 5 I 2 R = 0 or I 2 = Δ V 1 + Δ V 2 5 R = 12 V + 6 V 5(100Ω) = 18 V 500Ω = 0 . 036 A. (f) Solve for I 1 : Solve Loop 1 equation for I 1 in terms of I 2 , I 1 = Δ V 1 - I 2 R s R = 12 V - (0 . 036 A )(200Ω) 100Ω = 0 . 048 A. (g) Solve for I 3 : Use the junction equation to solve for I 3 , I 3 = I 2 - I 1 = (0 . 036 A ) - (0 . 048 A ) = - 0 . 012 A. (h) Check: Substitute the currents into one of the loop equations as a check. Use Loop 1 12 V - (0 . 048 A )(100Ω) - (0 . 036 A )(200Ω) = 0 , which does indeed add up, so we did it right. It is so easy to make a math error in a Kirchho±’s Law problem, so checking it is a necessity.
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_012 - Solution for Homework 13 RC and Review for Exam 2...

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