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PHYS 241-1 Test 3 Solutions: Loop Rule, Cramer's Rule, Capacitor Charging - Prof. David F., Exams of Physics

The solutions to test 3 of phys 241-1, including the application of kirchhoff's loop rule, cramer's rule, and the calculation of capacitor charging. It covers topics such as electric potential, capacitance, resistance, and time constants.

What you will learn

  • How do you find the electric potential at a certain point in an electrical circuit using the given information?
  • What are the solutions for the loop rules problem in PHYS 241-1 Test 3?
  • What is the charge stored on a capacitor in a given circuit, and how is it calculated?

Typology: Exams

2017/2018

Uploaded on 12/17/2018

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PHYS 241-1 Test 3 Key 10/18/18 p. 1
1. a. Use Kirchhoff’s loop rule. For loop 1 start at the positive terminal of the 6.00 v battery and proceed
clockwise around the loop. For loop 2 start at the positive terminal of the 12.0 v battery and proceed
counterclockwise around the loop. (There are other correct ways to write the loop rule for these loops.)
1 1 2 1 1 1 2
2 1 2 2 1 2
11
Loop 1: 4 10 10 5 6 2 0 21 10 6
Loop 2: 10 10 12 2 0 10 13 12
Using Cramer's rule:
21 10 ; 21 13 10 10; 173
10 13
6 10
1 6 13 10 12
;
12 13
173 1
I I I I I or I I
I I I I or I I
II

 1
2 2 2
; 0.243 A (10 points)
73
21 6
1 21 12 6 10
; ; 1.11 A (10 points)
10 12
173 173
I
I I I

b. Let point a be the ground below resistor R3 and point b be the junction above resistor R3. The electric
potential at point b is
a 1 2 b 1
a b 1
b
+10 10 Note that we treat as if its direction is correct.
Since 0 10 0.243 10 1.11 The negative value of makes the direction correction for us.
So 8.67 V . (15 points)
V I I V I
V V I
V
c. Since a fully-charged capacitor behaves is an open circuit, the potential difference across the capacitor is
the same as the potential difference across V2, which is the battery’s terminal voltage (not its emf). Start at
junction c on the right side of battery V2 and add voltage changes from junction c to junction b on the left
side of the battery:
c 2 b b c b c
12 2 ; 12 2 1.11 ; 9.78 V
This is the potential difference across the capacitor: 9.78 V
The charge stored on the capacitor is
: 10.0 F 9.78 V ; 97.8 C (10 points)
V I V V V V V
V
q CV q q
2. a.
0
Loop Rule: 0 0 or 0
IC: 0 (10 points)
q q dq q
iR i
C RC dt RC
qq
b.
00
1 1 1
From part a: ; ; ln ln
(where is a constant). Then . When 0, so that .
tt
RC RC
dq dq
dt dt q t k
q RC q RC RC
k q ke t q q q q e


(12 points)
The negative answer for
I1 indicates that this
loop current flows
opposite to the direction
assumed. (3 points)
I2
pf3

Partial preview of the text

Download PHYS 241-1 Test 3 Solutions: Loop Rule, Cramer's Rule, Capacitor Charging - Prof. David F. and more Exams Physics in PDF only on Docsity!

  1. a. Use Kirchhoff’s loop rule. For loop 1 start at the positive terminal of the 6.00 v battery and proceed

clockwise around the loop. For loop 2 start at the positive terminal of the 12.0 v battery and proceed

counterclockwise around the loop. (There are other correct ways to write the loop rule for these loops.)

1 1 2 1 1 1 2

2 1 2 2 1 2

1 1

Loop 1: 4 10 10 5 6 2 0 21 10 6

Loop 2: 10 10 12 2 0 10 13 12

Using Cramer's rule:

I I I I I or I I

I I I I or I I

I I

2 2 2

; 0.243 A (10 points) 73

; ; 1.11 A (10 points) 173 10 12 173

I

I I I

b. Let point a be the ground below resistor R3 and point b be the junction above resistor R3. The electric

potential at point b is

a 1 2 b 1

a b 1

b

+10 10 Note that we treat as if its direction is correct.

Since 0 10 0.243 10 1.11 The negative value of makes the direction correction for us.

So 8.67 V. (15 points)

V I I V I

V V I

V

c. Since a fully-charged capacitor behaves is an open circuit, the potential difference across the capacitor is

the same as the potential difference across V2, which is the battery’s terminal voltage ( not its emf). Start at

junction c on the right side of battery V2 and add voltage changes from junction c to junction b on the left

side of the battery:

c 12 2 2 b ;^ b c 12 2 1.11 ;^ b c 9.78 V

This is the potential difference across the capacitor: 9.78 V

The charge stored on the capacitor is

: 10.0 F 9.78 V ; 97.8 C (10 points)

V I V V V V V

V

q CV q q

  1. a.

Loop Rule: 0 0 or 0

IC: 0 (10 points)

q q dq q iR i C RC dt RC

q q

b.

0 0

From part a: ; ; ln ln

(where is a constant). Then. When 0, so that.

t t RC RC

dq dq dt dt q t k q RC q RC RC

k q ke t q q q q e

 

(12 points)

The negative answer for

I 1 indicates that this

loop current flows

opposite to the direction

assumed. (3 points)

I 2

2 c.

0 0 0 ;^ (10 points)

The answer is negative because the current flows opposite to the direction assumed.

t t t RC RC RC

dq d dq q q q e e i e dt dt dt RC RC

 ^  ^ 

d.

The time constant of this circuit is  RC.

1 When the current in the circuit and the charge on the capacitor are 37% (0.368 or 36.8%)

of their initial values. (10 points)

t e

   

3. For the segment shown in Figure 3:    

0 2

, where

1

dx dR x x A (^) bx

. The total resistance is found

by integrating from 0 to L :

      ^ 

0

0 2

2 2 2 2

0 0 0

0

Let 1. Then :

1

1 (10 points) 1 1 1

L

L

R dx u bx du bdx A (^) bx

bdx du dx dx

bx b^ bx b^ u^ bx b^ bx

L

R or R or R A b bx Ab bL A bL

 ^  ^  ^ 

 ^   ^  

Continued on the next page

q

i

dx

x

x

Figure 3

y (^) This segment has resistance dR ,

resistivity ( x ), length dx and

cross-sectional area A.

0 L