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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
Typology: Exams
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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
2 2 2
; 0.243 A (10 points) 73
; ; 1.11 A (10 points) 173 10 12 173
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)
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)
q CV q q
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)
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
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
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.