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PHY2049
Spring 2006
Exam 2 Solutions
Prof. Darin Acosta
Prof. Greg Stewart
March 27, 2006
Exam 2 Solutions
a
______
_____________
b
____________
__________

5
Ω
>i
1

<
i
2
5
Ω

⎯
3V
10
Ω
⎯
6 V
⎯





↓
i
3

⎯
d
___________________________
c
___________
_____________
10
Ω
1.
In the multiloop circuit shown, what is the current in the left hand loop (abcd), in
units of Amperes?
a.
3/55
b.
3/11
c.
12/55
d.
0
e.
9/55
Solution:
Use Kirchhoff's rules (sum of currents at a point = 0 and sum of EMF's
around a loop = 0) to obtain:
(direction of currents chosen  remember these can be
180
o
wrong, which will give a minus sign in the solution  are shown in
red
)
eq. #1: i
1
+ i
2
= i
3
(sum of currents at point b)
eq. #2: 5i
1
 10i
3
+3V=0 (sum of EMF's around left hand loop starting at point a
and going
clockwise)
eq. #3:
+5i
2
 6V +10i
2
+ 10i
3
= 0 (sum of EMF's around right hand loop starting at
point b and going clockwise)
Solve these 3 eqns for i
1
:
#3
⇒
15 i
2
+ 10i
3
=6
#2
⇒
10 i
3
= 5i
1
+ 3
put this in eq. above
* 15i
2
+ (5i
1
+ 3) = 6
also, put eq. above as i
3
= (1/10)(5i
1
+ 3) in eq. #1 to get
** i
1
+ i
2
= (1/10)(5i
1
+ 3)
use * to express i
2
= (6 3 +5i
1
)/15 and put in ** to get
i
1
+ 1/5 + (1/3) i
1
= (1/2)i
1
+ 3/10
solve for i
1
, get i
1
= 3/55 Amp
However, the wording of the problem was found to be ambiguous, as the current
through the middle branch (10 ohm resistor) is not the same as the left current
(through the 3V battery). Both were marked correct, so if one chose i
3
=3/11 that is
also correc
t.
1
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View Full DocumentPHY2049
Spring 2006
Exam 2 Solutions
⎯⎯⎯⎯
/
⎯⎯⎯



⎯
q
0

____________________
2.
A capacitor starts fully charged with charge q
0
.
It is then, through the closing of a
switch, discharged through a series resistor, see circuit in Figure.
If C=10
μ
F and
R=100
Ω
, how long in units of milliseconds (1 millisecond = 10
3
s) does it take the
charge on the capacitor to reach 0.5 q
0
?
(1
μ
F = 10
6
F)
a.
0.69
b.
0.63
c.
0.37
d.
0.31
e.
0.5
Solution:
q = q
0
e
t/RC
if want q/q
0
= 0.5, then ln(0.5) = t/RC,
or 0.6931 = t/(100* 10
5
) or t=0.69 10
3
s, so answer (in units of milliseconds) is 0.69.
3. An electron is accelerated from rest by a potential difference of 20,000V.
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