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Physics 204 – Section 9
QUIZ 3
17 February 2013
Name:
1. (10 points) Find the current I1 through the 5 W resistor and the current I2 through
the 10 W resistor.
Solution: From Kirchoff’s second law, moving around the bottom loop we get
12 V = I3 (2 W) + I2 (10 W).
(1)
Moving around the top loop,
I1 (5 W) = I2 (10 W).
(2)
I3 = I1 + I2 .
(3)
From Kirchoff’s first law,
From equation (1),
I3 =
12 V
I2 (10 W)
.
2W
(4)
Plugging this in to (3)
12 V
I2 (10 W)
12 V I2 (10 W)
= I1 + I2 =) I1 =
2W
2W
I2 .
(5)
Then plugging this into (2) gives
✓
◆
(6)
(12 V)(5 W)/(2 W)
= 0.75 A
10 W + 5 W + (10 W)(5 W)/(2 W)
(7)
12 V
I2 (10 W)
2W
I2 (5 W) = I2 (10 W),
which implies
I2 =
and consequently,
I1 =
12 V
(0.75 A)(10 W)
2W
0.75 A = 1.5 A.
(8)
Physics 204 – Section 11
QUIZ 3
17 February 2013
Name:
1. (10 points) Find the current I1 through the 5 W resistor and the current I2 through
the 10 W resistor.
Solution: From Kirchoff’s second law, moving around the left loop we get
12 V = I1 (5 W) + I2 (10 W).
(1)
Moving around the loop on the right,
9 V = I2 (10 W).
From equation (2),
I2 =
9V
= 0.9 A.
10 W
(2)
(3)
From (1),
I1 =
12 V
I2 (10 W)
12 V
=
5W
(0.9 A)(10 W)
= 0.6 A.
5W
(4)
Physics 204 – Section 13
QUIZ 3
17 February 2013
Name:
1. (10 total points) An RC circuit contains a capacitor of C = 1 ⇥ 10 3 F, a resistor of
resistance R, and a battery that applies an emf of 12 V. It takes 0.9 s for the capacitor
to gain a charge of 0.01 C on each plate.
(a) (3 points) If the circuit is left to charge for long enough, what will the charge
and each plate become?
Solution: The charge stored by a capacitor is given by
Q0 = CV0 = (1 ⇥ 10
3
F)(12 V) = 0.012 C.
(1)
(b) (7 points) Find the value of R.
Solution: The charge on the plate as a function of times is
Q ( t ) = Q0 (1
e
t/RC
).
(2)
If we leave the term with the exponential on one side and move everything
else to the other, then
Q0 Q = Q0 e t/RC .
(3)
Taking the natural logarithm of both sides gives
ln( Q0
Q) = ln( Q0 e
t/RC
) = ln Q0 + ln e
t/RC
= ln Q0
t
.
RC
(4)
Now we can solve for R:
R =
C [ln( Q0
t
Q)
ln Q0 ]
0.9 s
=
(1 ⇥ 10 3 F)[ln(0.012 C 0.01 C)
= 502.3 W.
(5)
ln(0.012 C)]
(6)
(7)
Physics 204 – Section 20
QUIZ 3
17 February 2013
Name:
1. (a) (3 points) What is the magnitude of the magnetic field a distance 0.5 m away
from a long wire carrying 2 A of current?
Solution: The magnitude of the magnetic field from a long wire
B=
µ0 I
(4⇡ ⇥ 10 7 T m/A)(2 A)
=
= 8 ⇥ 10
2⇡r
2⇡(0.5 m)
7
T.
(1)
(b) (3 points) What is the magnitude of the magnetic field a distance 0.5 m away
from a wire carrying 3 A of current?
Solution: The magnitude of the magnetic field from a long wire
B=
µ0 I
(4⇡ ⇥ 10 7 T m/A)(3 A)
=
= 1.2 ⇥ 10
2⇡r
2⇡(0.5 m)
6
T.
(2)
(c) (4 points) There are two long wires, separated by 1 m. One carries 2 A of current
and the other caries 3 A of current in the same direction as the first wire. Find
the magnitude of the net magnetic field half way between the two wires.
Solution: A point that is half way between the wires is 0.5 m from each one.
Therefore, the magnetic field from each wire is the same as we found in parts
(a) and (b). Assume that the wires are running from left to right with the
currents flowing toward the right. Additionally, assume that the wire with
2 A is behind the wire with 3 A. Then the 8 ⇥ 10 7 T magnetic field from the
2 A wire is pointing down. Additionally, the 1.2 ⇥ 10 6 T magnetic field from
the 3 A wire is pointing up. The net magnetic field is the sum of those two
magnetic fields, so the magnitude is 1.2 ⇥ 10 6 T 8 ⇥ 10 7 T = 4 ⇥ 10 7 T.