HW12 - 12th HW PDF

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Solution for Homework 12 Kirchhoff’s Laws Solution to Homework Problem 12.1(Reduce Resistance Network) Problem: For the circuit below, R1 = 3.0Ω, R2 = 6.0Ω, R3 = 6.0Ω, R4 = 3.0Ω, and R5 = 3.0Ω, and ∆V0 = 9.0V. To receive any IBC for this problem, your symbolic work must clearly express your method. (a)What is the equivalent resistance of the network? (b)What is the voltage across R5 ? (c)What is the current through R1 ? (d)How much power is dissipated in R3 ?

R1

R2

R3 R5

R4

∆V0 Solution to Part (a) Find the Equivalent Resistance: The resistors R1 and R2 are in series and have equivalent resistance R12 = R1 + R2 = 3Ω + 6Ω = 9Ω. The resistors R4 and R5 are also in series and have equivalent resistance R45 = R4 + R5 = 3Ω + 3Ω = 6Ω.The combination R12 , R3 , and R45 is in parallel and can be reduced to give the equivalent resistance, 1 1 1 1 1 1 1 + + = + + = 9Ω 6Ω 6Ω R12345 R1 R2 R3 giving

9 R12345 = Req = Ω 4 Grading Key: Solution to Part (a) 4 Points Solution to Part (b)

The full battery voltage is across R45 so the current through R45 is given by Ohm’s Law, 3 I = ∆V0 /R45 = 9V/6Ω = A 2

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This current flows through R5 because it is in series.Therefore the voltage across R5 is 3 9 ∆V5 = I45 R5 = ( A)(3Ω) = V 2 2 This can also be obtained by observing the resistor are of equal value so the voltage must split equally between them. Grading Key: Solution to Part (b) 4 Points Solution to Part (c) The full battery voltage is across R12 so the current through R12 is given by Ohm’s Law, I = ∆V0 /R12 = 9V/9Ω = 1.0A This current flows through R1 because it is in series. Grading Key: Solution to Part (c) 4 Points Solution to Part (d) The voltage across R3 is the battery voltage, so the power is P = I∆V =

(∆V )2 27 (9V)2 = W = 6Ω 2 R

Grading Key: Solution to Part (d) 3 Points Total Points for Problem: 15 Points

Solution to Homework Problem 12.2(Fancy Kirchhoff Problem) Problem: For the resistance network below, (a)Which if any of the resistors below are in series? There may be more than one or no series combination. (b)Which if any of the resistors below are in parallel? There may be more than one or no parallel combination. (c)Write a loop equation for loop 1 (KCDEJK). (d)Write a loop equation for loop 2 (HJEFGH). (e)Write a junction equation for junction J. (f)Write a junction equation for junction E. R3 I11 I5 R6

I3

I6 R4 R7

R1

I4

R9

R5 I2

I8

I7 R2

2

I1

Solution to Part(a) Series R4 and R6 Grading Key: Part (a) 2 Points Solution to Part(b) Parallel R1 and R7 and R2 and R9 . Grading Key: Part (b) 2 Points Solution to Part(c) I4 R5 + 6V − 1.5V + I3 R6 + I3 R4 = 0 Grading Key: Part (c) 5 Points Solution to Part(d) 3V − I3 R4 − I3 R6 + I11 R3 − I1 R2 = 0 Grading Key: Part (d) 5 Points Solution to Part(e) I7 + I2 = I3 Grading Key: Part (e) 2 Points Solution to Part(f) I5 + I11 + I3 = 0 Grading Key: Part (f) 2 Points Total Points for Problem: 18 Points

Solution to Homework Problem 12.3(Two Loop Circuit Giving Current) Problem: In this resistance network, R1 = 5.00Ω, R2 = 10.0Ω, ∆V1 = 1.00V and ∆V2 = 5.00V . Simplify ALL simple series and parallel combinations to their equivalent resistance. Draw the simplified circuit. Write the junction equation for junction b and loop equations for Loop 1 and Loop 2. Solve for the currents. Since I’m feeling nice today, I’ll give you one of the currents: I1 = −0.514A. Solve for I2 and I3 . Make sure your calculation is well ordered and labelled so we can follow it.

R1 I1 a

I3

R2

R2

b ∆V2

R1

∆V1 Loop 2

Loop 1 I2

f

3

c

R1

R2

R1

e

d

Solution (a) Reduce Series and Parallel Combination: First, we simplify the 3 resistors in parallel. 1 1 1 1 1 = = + + 2Ω Rp 5Ω 5Ω 10Ω

I1

I3

a

c

b Rp

Rp = 2Ω Now simplify the resistors in series. ∆V1

Rs = R2 + R1 + R2 = 10Ω + 5Ω + 10Ω = 25Ω

Rs

∆V2 Loop 2

Loop 1 R1

I2

f

e

d

(b) Write Junction Equation: Now write the junction equation for b. I1 = I2 + I3 (c) Write Loop Equations: Write both loop equations. First, write Loop 1 and substitute the voltages: ∆Vf a + ∆Vab + ∆Vbe + ∆Vef = 0 ∆V1 − I1 Rp − ∆V2 − I2 R1 + 0 = 0 1V − (2Ω)I1 − 5V − (5Ω)I2 + 0 = 0 Now do the same for Loop 2: ∆Veb + ∆Vbc + ∆Vcd + ∆Vde = 0 I2 R1 + ∆V2 − I3 Rs + 0 = 0 (5Ω)I2 + 5V − (25Ω)I3 + 0 = 0 (d) Solve: Now one would proceed with the unenviable task of solving this system of equations, but since I gave you I1 you can just breeze through it. Plug this into Loop 1 equation and find I2 . 1V − (−0.514A)(2Ω) − 5V − I2 (5Ω) = 0 (0.514A)(2Ω) − I2 (5Ω) = 4V I2 = −0.595A Now use the junction equation to find I3 . I1 = I2 + I3 I3 = I1 − I2 (−0.514A) − (−0.595A) = 0.081A (e) Check: Check Loop 1 1V − (2Ω)(−0.514A) − 5V − (5Ω)(−0.595A) + 0 = 0 Check Loop 2 (5Ω)(−0.595A) + 5V − (25Ω)(0.081A) + 0 = 0 2 point(s) : Reducing Series and Parallel Combinations 1 point(s) : Redraw Circuit 3 point(s) : Junction Equation 4 point(s) (2 times) : Loop Equations 2 point(s) : Solving For Currents Correctly Total Points for Problem: 16 Points

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