Onlinehomework-12-sln PDF

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MC HW 12...

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Solution for Online Homework 12 Kirchhoff’s Laws Solution to Online Homework Problem 12.1(Three Bulbs in Parallel) Problem: Identical light bulbs B1 and B2 are in parallel. How does the brightness of B2 change as a third identical light bulb is added in parallel to B2 ? Assume a perfect battery. Select One of the Following: (a) The brightness of B2 increases. (b) The brightness of B2 decreases. (c-Answer) The brightness of B2 stays the same. Solution If the battery is perfect, adding another bulb in parallel does not affect the other bulbs because it does not change the potential difference. Since it does not change the potential difference (and resistance) of the other bulbs, the current through the other bulbs will not change. Brightness is our qualitative method of measuring current. Because the current doesn’t change when the third bulb is added, the brightness of B1 or B2 is not changed.

2 point(s) : Brightness does not change. 2 point(s) : Drawing of circuit. Total Points for Problem: 4 Points

Solution to Online Homework Problem 12.2(What Happens to Current as it Passes Through a Resistor?) Problem: As current passes through a resistor, which of the following is true? Select One of the Following: (a) The current has less energy before passing through the resistor than after and the potential increases. (b-Answer) The current loses energy as it passes through the resistor and the potential is decreased. (c) The current is not affected and the potential remains the same. (d) The current does not pass through the resistor, it get stuck in the resistor. (e) Charge is converted into heat in the resistor. Solution The current loses potential potential energy and potential decreases. Total Points for Problem: 3 Points

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Solution to Online Homework Problem 12.3(Zero Current Implies Zero Potential Difference Across Resistor) Problem: The two ends of a resistor with resistance R are connected to point A and point B in a circuit. Both ends of the resistor are held at a potential V . What is the current through the resistor? Select One of the Following: (a) V/R (b-Answer) 0 (c) Cannot be determined unless the details of the full circuit are known. Solution With B closed and A open, no current flows through the resistor, so the potential difference is zero. Total Points for Problem: 3 Points

Solution to Online Homework Problem 12.4(Kirchhoff’s Law with Series Element) Problem: Analyze the circuit to the right with ∆V1 = 12V , ∆V2 = 6V , R1 = R2 = R3 = 10Ω. Use the directions for the currents drawn on the diagram. Write a junction equation for junction j .

R1 I1 ∆V1

I2

R3

∆V2 I3

j

R2

Loop Select One of the Following: (a) I2 = I3 (b) I1 + I3 = I2 (c) I2 + I3 = I1 (d-Answer) I1 + I2 = I3 (e) 0 = I1 + I2 + I3 Solution Introduce Currents and Write Junction Equation: Using the current directions given in the figure above, if we conserve current at node j, we find: I1 + I2 = I3 P Current into the junction is positive, and current out of the junction is negative for Ii = 0. Total Points for Problem: 3 Points

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Solution to Online Homework Problem 12.5(Compute Potential at Point from Reference and Potential Difference - Circuits) Problem: In the circuit to the right, let the point A be defined to be at zero potential. The battery is a 12V battery. The potential drop as the current I passes across the left resistor is −3V and potential drop across the right resistor is −9V. What is the potential at the point C between the resistors?

_

Select One of the Following:

A

(a) 0V

+

I

(b-Answer) 3V (c) 6V

C

D

(d) 9V

B

(e) 12V

Solution The potential at C is the potential at the reference plus the changes: 0V + 12V + −9V = 3V. Total Points for Problem: 3 Points

Solution to Online Homework Problem 12.6(Voltage Across Parallel Resistor) Problem: In the resistance network to the right the resistors are R1 = 2Ω, R2 = 4Ω, R3 = 4Ω, and R4 = 2Ω. The battery has voltage 9V. Calculate the current through R2 .

R1

Select One of the Following: (a) 1.5A (b) 0.83A

R2

(c) 1.33A (d-Answer) 0.75A (e) 0.50A

R4

Solution

3

R3

The equivalent resistance of the parallel circuit is 1 1 1 + = R2 R3 Rp or Rp = 2Ω. The equivalent circuit with the parallel resistor reduced is drawn to the right. The equivalent resistance of the full circuit is then Req = R1 + Rp + R4 = 6Ω. The current drawn by the circuit is then I = ∆V/Req = 32 A. The potential difference across Rp is then ∆Vp = IRp = 3V. This is the same potential difference that exists across R2 because they are in parallel. Since ∆V2 = 3V and R2 = 4Ω, by Ohm’s Law I2 = ∆V2 /R2 = 3V/4Ω = 3/4A

R1

Rp

R4

Total Points for Problem: 3 Points

Solution to Online Homework Problem 12.7(Parallel Elements Identical - All Have Equal Current) Problem: The figure to the right shows an electric circuit where all the resistors are identical R1 = R2 = R3 = R. How does the current, I2 , flowing in R2 compare to the current, I3 , flowing in R3 ?

R1

Select One of the Following:

R2

(a) I2 > I3

R3

(b) I2 < I3 (c-Answer) I2 = I3

Solution R1 has the most current, since the current splits after R1 as it flows into the parallel combination R2 and R3 . Because R2 and R3 are identical and connected in parallel (same potential difference), they must have the same amount of current flowing through each of them (Ohm’s Law: ∆V = IR). Total Points for Problem: 3 Points

Solution to Online Homework Problem 12.8(Electric Potential of a Circuit Drops Across a Resistor) Problem: How does the electric potential of an electric current change as it passes through a resistor? Select One of the Following: (a-Answer) The current has less electric potential after passing through the resistor. (b) The current has more electric potential after passing through the resistor. (c) The current has the same electric potential after passing through the resistor. (d) The current may either gain or lose electric potential after passing through the resistor. Solution 4

The electric potential of the current decreases as it passes through a resistor. Total Points for Problem: 3 Points

Solution to Online Homework Problem 12.9(What Is a Perfect Battery) Problem: What would represent a perfect battery? Select One of the Following: (a) A battery that delivered the same resistance at all voltages. (b) A battery that delivered the same resistance at all currents. (c-Answer) A battery that delivered the same potential difference at all currents. (d) A battery that delivered the same current at all voltages. Solution A perfect battery has no internal resistance, so it delivers the same potential difference at all currents. Total Points for Problem: 3 Points

Solution to Online Homework Problem 12.10(Ranking Resistors by the Amount of Current Passing Through Them) Problem: The figure to the right shows an arrangement of identical resistors in a circuit. With the switch open, a current I flows in the circuit. How much current flows in the circuit if the switch is closed?

R2

R3

Select One of the Following: (a) 0 (b) I (c) I/2

S

(d) I/4

R4

R1

(e-Answer) 2I (f) 4I

+ _

Solution Before the switch is closed, there are four resistors in series for an equivalent resistance of 4R.If the switch is closed, all current goes through the switch and none through resistors R2 and R3 .Therefore, these two resistors can be removed from the circuit.This leaves two resistors in series R1 = R2 for an equivalent resistance of Req = 2R. Since the resistance is halved, the current is doubled: ∆Vbatt = IReq . Total Points for Problem: 6 Points

Solution to Online Homework Problem 12.11(Kirchhoff’s Law with Series Element)

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Problem: Analyze the circuit to the right with ∆V1 = 12V , ∆V2 = 6V , R1 = R2 = R3 = R4 = 10Ω. Use the directions for the currents drawn on the diagram. Write the loop equation for the loop drawn on the diagram.

R1

Select One of the Following: (a) ∆V1 + I1 R1 + I2 R3 + ∆V2 − I3 R2 + I1 R4 = 0

∆V1 I2

(b) ∆V1 − I1 R1 − I2 R3 − ∆V2 + I3 R2 − I1 R4 = 0 (c) ∆V1 + I1 R1 + ∆V2 + I3 R2 + I1 R4 = 0

R3

∆V2 I3

I1

(d) −∆V1 − I1 R1 − ∆V2 − I3 R2 − I1 R4 = 0

R4

(e-Answer) −∆V1 − I1 R1 − ∆V2 + I3 R2 − I1 R4 = 0 (f) −∆V1 + I1 R1 + ∆V2 − I3 R2 + I1 R4 = 0

j

R2

Loop

(g) −∆V1 − I1 R1 − I2 R3 − ∆V2 + I3 R2 − I1 R4 = 0 Solution Write Loop Equations: Kirchhoff’s Loop equation is given by ∆V = 0 Σ

Use the directions of the loop and of the current drawn in the diagram to determine if the potential increases or decreases across each battery and resistor. (1) The potential drops across the first battery because the loop goes from positive to negative, so −∆V1 . (2) Ohm’s Law says the the change in potential across a resistor R with current I is ∆V = IR At resistor R1 , the loop is to the right, and the current I1 is to the right, so there is a decrease in potential, −I1 R1 . (3) The potential again drops across the second battery because the loop goes from positive to negative, −∆V2 . (4) Again using Ohm’s Law, the loop is to the left at resistor R2 , but the current I3 is to the right, so there is a rise in potential, +I3 R2 . (5) At R4 , the current I1 correlates with the loop, so there is a drop in potential of −I1 R4 . Add the changes for each piece discussed to get the total change in potential: −∆V1 − I1 R1 − ∆V2 + I3 R2 − I1 R4 = 0 Total Points for Problem: 3 Points

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