Download Circuits Review 2007-2008

Circuits Review Sheet Answers
AP Physics
1. a) What is the current when a 15 V battery is connected across a 120 k resistor?
V
I=R=
15 V
= 1.25 x 10-4 A
120000 
b) How many electrons pass through the resistor each second?
1 eI = 1.25 x 10-4 A x 1.609 x 10-19 C = 7.81 x 1014 e–
2. An electric thermometer is built with a 5.5 V battery, a 1 cm long cylindrical nichrome wire and an
ammeter in a spiffy plastic case. The reading for current is then translated into a temperature reading.
a) When the temperature is 20 °C, the current is 2.54 A. What is the resistance at that temperature?
V 5.5 V
R = I = 2.54 A = 2.17 
b) How thick (diameter) is the nichrome wire?
l
R =A
l
.01m
A = R = 100 x 10-8 W·m
= 4.61 x 10-9 m2
2.17 
A = r2
A
r=
= 1.47 x 10-9 m2 = 3.83 x 10-5 m

d = 2 r = 7.66 x 10-5 m
c) A person puts the thermometer in their mouth, which is at 37 °C. What is the resistance of the wire
now?
 = 0(1 + ·T)
 = 100 x 10-8 W·m (1 + 0.40 x 10-3 x (37 °C – 20 °C)) = 100.68 x 10-8 ·m
l
R =A
.01m
R = 100.68 x 10-8 W·m x 4.61 x 10-9 m2 = 2.18 
or
l
l
l
R =A = 0(1 + ·T)A = 0A(1 + ·T) = R0(1 + ·T)
R = 2.17  (1 + 0.40 x 10-3 x (37 °C – 20 °C)) = 2.18 
d) How much current flows through the resistor while it is in the person’s mouth?
V 5.5 V
I=R=
= 2.52 A
2.18 
e) How about that plastic case?
It’s spiffy!
3. (hard) A resistor connected to a constant voltage supply allows 0.l0417 A through at 0 °C and
0.0332 A at 500 °C. What material is the resistor made from?
(notice that the numbers are changed from what was passed out during class)
V
V
V
R = I so RH = I and RC = I
H
C
V
RH IH IC 0.10417 A
RC = V = IH = 0.0332 A = 3.14
IC
RC = R0(1 + ·TC) and RH = R0(1 + ·TH)
RH R0(1 + ·TH)
RC = R0(1 + ·TC) = 3.14
R0(1 + ·TH) = 3.14 R0(1 + ·TC)
(1 + ·TH) = 3.14 (1 + ·TC)
1 + ·TH = 3.14 + 3.14·TC
·TH – 3.14·TC = 3.14 – 1
TH – 3.14·TC) = 2.14
[(500°C – 20°C) – 3.14·(0°C – 20°C)] = 2.14
[542.8°C] = 2.14
 = 0.00394
and that must be platinum!
For numbers given in the paper handed out in class:
V
RH IH IC 0.10417 A
RC = V = IH = 0.0348 A = 2.99
IC
RH = R0(1 + ·TH)
RH
= (1 + ·TH) = 2.99
R0
·TH = 1.99
·(500°C – 20°C) = 1.99
·(480°C) = 1.99
 = 0.0041 and that is between platinum and aluminum .
4. A copper wire (at 20 °C), which is 1 mm thick and 30 cm long is connected to a 1.5 V battery.
a) How much current flows through the wire?
l
0.30 m
R = = 1.70 x 10-8 m
= 0.0065 
A
7.85 x 10-7 m2
A = r2 = 3.14159 x (.0005 m)2 = 7.85 x 10-7 m2
V
I=R=
1.5 V
= 231 A
0.0065 
b) What is the power use in the wire?
P = VI = 1.5 V x 231 A = 346.5 W
c) (hard) If all of the power goes into heating up the wire (and copper has a density of 8.9 g/cm3 and a
specific heat of 0.39 J/g·°C, how hot will it be after 10s?
Q = E = Pt = 346.5 W x 10s = 3465 J
Q = mcT so
Q
3465 J
T = mc = 2.1004 g x 0.39 J/g·°C = 4230 °C
T = T0 + T = 20 °C + 4230 °C = 4250 °C!
(of course, in reality it would give some of the energy away once it got hotter than the air)
m =  x V where  stands for density
m = 8.9 g/cm3 x 0.236 cm3 = 2.1004 g
V = r2l for cylinder
V = 3.14159 x (0.05 cm)2 x 30 cm = 0.236 cm3
5. A car compartment light bulb has a power rating of 15 W when connected to a 12 V battery.
a) What is the resistance of the light bulb?
P=
V2
so
R
V2 (12 V)2
R = P = 15 W = 9.6 
b) How much current passes through the light bulb?
V 12 V
I=R=
= 1.25 A
9.6 
6. Your space heater has a tungsten coil with a cold (20°C) resistance of 5.36  connected to a 120 V
outlet.
a) What is the current through the cold heater?
V 120 V
I=R=
= 22.4 A
5.36 
b) What is the power consumption when it is first turned on?
P = IV = 22.4 A x 120 V = 2688 W
c) What is the resistance when heater is warmed up to 400 °C?
R = R0(1 + T) = 5.36  (1 + 4.5 x 10-3 x (400°C – 20 °C)) = 14.5 
d) What is power consumption of warm heater?
V 120 V
I=R=
= 1.38 A
14.5 
P = IV = 1.38 A x 120 V =166 W
7. Find the current through each resistor and the voltage across each resistor in this
circuit:
RT = R1+R2+R3+R4 = 60  + 15  + 10  + 25  = 110 
V
8V
IT = R =
= 0.073 A
110 
T
Vn = In·Rn for any section of circuit labeled n.
For series circuit IT = I1 = I2 = I3 = I4 = 0.073 A
check that voltages add up to total
V1 = I1·R1 = 0.073 A · 60 = 4.38 V
For series circuit VT = V1 + V2 + V3 + V4
V2 = I2·R2 = 0.073 A · 15 = 1.09 V
VT = 4.38 V + 1.09 V + 0.73 V + 1.82 V = 8.02 V
V3 = I3·R3 = 0.073 A · 10 = 0.73 V
V4 = I4·R4 = 0.073 A · 25 = 1.82 V
8. Find the current through each resistor and the voltage across each resistor in this circuit:
1
1
1
1
1
1
1
1
1
RT = R1 + R2 + R3 + R4 = 8  + 4  + 6  + 12 
1
3
6
4
2
15
24 
=
+
+
+
=
R
T=
RT 24  24  24  24  24 
15 = 1.6 
V 32 V
IT = R =
= 20 A
1.6 
T
Vn
and In = R for any circuit element n
n
In parallel circuit: VT = V1 = V2 = V3 = V4 = 32 V
V1 32 V
V2 32 V
V3 32 V
I1 = R =
=4A
I2 = R =
=8A
I3 = R =
= 5.33 A
8
4
6
1
2
3
Check that currents add up:
IT = I1 + I2 + I3 + I4 = 4 A + 8 A + 5.33 A + 2.67 A = 20 A
V4 32 V
I4 = R =
= 2.67 A
12 
4
9. Find the current through each resistor and the voltage across each resistor in this circuit:
RT = R1 + RA + RB = 10  + 36  + 15  = 61 
1
1
1
1
1
RA:
= +
=
+
RA R2 RS 90  60 
1
2
3
5
RA = 180  + 180  = 180 
180 
RA = 5 = 36 
RS = R2 + R3 = 36  + 24  = 60 
1
1
1
1
1
3
1
4
60 
RB: R = R + R =
+
=
+
=
RB = 4 = 15 
20  60  60  60  60 
B
5
6
V 100 V
IT = R =
= 1.64 A
61 
T
Vn = In·Rn for any section of circuit labeled n.
For series portion IT = I1 = IA = IB = 1.64 A
V1 = I1·R1 = 1.64 A · 10 = 16.4 V VA = IA·RA = 1.64 A · 36 = 59.0
V
VB = IB·RB = 1.64 A · 15 = 24.6 V
check that voltages add up: VT = V1 + VA + VB = 16.4 V + 59.0 V + 24.6 V = 100 V
Vn
and In = R for any circuit element n
n
In parallel portion: VA = V2 = VS = 59 V and VB = V5 = V6 = 32 V
V2 59 V
VS 59 V
A: I2 = R =
= 0.66 A
IS = R =
= 0.98 A
90 
60 
2
S
check that currents add up: IT = I2 + IS = 0.66 A + 0.98 A = 1.64 A
V5 24.6 V
V6 24.6 V
B: I5 = R =
= 1.23 A
I6 = R =
= 0.41 A
20 
60 
5
6
check that currents add up: IT = I5 + I6 = 1.23 A + 0.41 A = 1.64 A
Finally, for series portion IS = I3 = I4 = 0.98 A
V3 = I3·R3 = 0.98 A · 36 = 35.3 V
V4 = I4·R4 = 0.98 A · 24 = 23.5 V
check that the voltages add up: VS = V3 + V4 = 35.3 A + 23.5 A = 58.8 A
10. Find the current through each resistor in this circuit:
Apply Kirchoff’s Rules:
Junction Rule:
A: I1 – I2 + I3 + I4 = 0
Loop Rule:
1: 27 V – 64·I2 – 8·I3 = 0
or
– 64·I2 – 8·I3 + 27 V = 0
2: + 4·I1 – 18 V + 64·I2 = 0 or
4·I1 + 64·I2 – 18 V = 0
3: + 8·I3 – 16·I4 = 0
4: – 18 V – 16·I4 + 27 V + 4·I1 = 0 or 4·I1 – 16·I4 + 9 V =
0
5: 4·I1 – 18 V – 8·I3 + 27 V = 0
or 4·I1 – 8·I3 + 9 V = 0
I need four equations to solve four unknowns. I chose the junction rule and the first three loop rules:
Put them into a matrix:
in row reduced echelon form: Which means:
1 -1 1 1 0
.776
I1 = -.776 A right or 0.776 A to the left
 0 -64 -8 0 27 
 10 01 00 00 -.330

I2 = 0.330 A down
 4 64 0 0 -18 
 0 0 1 0 -.737 
I3 = 0.737 A to the right
 0 0 8 -16 0 
 0 0 0 1 -.369 
I4 = 0.369 A to the right
11. Find the current through each resistor and the voltage across each resistor in this circuit:
Apply Kirchoff’s Rules:
Junction Rule:
A: I1 + I2 – I3 = 0
all other junctions end up giving no valuable information
Loop Rule:
1: – 13·I1 + 5·I2 = 0
2: – 5·I2 – 3·I3 – 20 V + 30 V = 0 or
– 5·I2 – 3·I3 + 10 V = 0
3: 20 V – 11·I5 – 7·I4 = 0
or
– 7·I4 – 11·I5 + 20 V = 0
4: – 10 V + 11·I5 = 0
or
11·I5 – 10 V = 0
5: – 13·I1 – 3·I3 – 10 V – 7·I4 + 30 V = 0 or – 13·I1 – 3·I3 – 7·I4 + 20 V =
0
6: 10 V – 20 V + 7·I4 = 0
or
7·I4 – 10 V = 0
and you can make a lot more loops if you put your mind to it…but most
of them are just combinations of the loops we already have (like loop 6 is
a combination of loops 3 and 4) and won’t help us.
I need five equations to solve five unknowns. I chose the junction rule and the first four loop rules:
Put them into a matrix:
in row reduced echelon form:
Which means:
1 1 -1 0 0 0
1 0 0 0 0 -.42
I2 = 1.09 A to the right
 -13 5 0 0 0 0 
 0 1 0 0 0 -1.09 
I3 = 1.51 A down
 0 -5 -3 0 0 10 
 0 0 1 0 0 -1.51 
I4 = 1.43 A up
 0 0 0 -7 -11 20 
 0 0 0 1 0 -1.43 
I5 = 0.91 A down
 0 0 0 0 11 -10 
 0 0 0 0 1 -.91 
I1 = 0.42 A to the right
12. A timing circuit is based on an RC circuit. The 5.2 mF capacitor is linked in series to a 8 k
resistor and attached to a 120 V power supply to charge it.
a) What is the time constant for this circuit?
 = RC = 5.2 mF x 8 k = 41.6 s
b) i) What is the voltage across the capacitor at the moment it begins charging?
VC = 0 when it is uncharged (Q = 0 and V = Q/C)
ii) What is the voltage across the resistor at the moment the capacitor begins charging? How much
current is flowing through the resistor at that point?
VB = VC + VR because the resistor and capacitor are in series
VR = VB – VC = 120 V – 0 V = 120 V
V 120 V
I=R=
= 0.015 A
8000 
c) i) What is the voltage across the capacitor 10 seconds after it begins charging?
-t
-10s




RC
41.6s
 = 120 V1 – e
 = 25.6 V
VC = V01 – e
( )
( )
ii) What is the voltage across the resistor 10 seconds after the capacitor begins charging? How
much current is flowing through the resistor at that point?
V 94.4 V
VR = VB – VC = 120 V – 25.6 V = 94.4 V
I=R=
= 0.0118 A
8000 
d) i) What is the voltage across the capacitor 100 seconds after it begins charging?
-t
-100s




RC
41.6s
 = 120 V1 – e
 = 109.2 V
VC = V01 – e
( )
( )
ii) What is the voltage across the resistor 100 seconds after the capacitor begins charging? How
much current is flowing through the resistor at that point?
VR = VB – VC = 120 V – 109.2 V = 10.8 V
V 10.8 V
I=R=
= 0.0014 A
8000 
13. The shock that a taser delivers come from a highly charged capacitor discharging through a
resistor (with a lot of other electronics to make it more nasty). Let’s presume that one taser uses a 900
mF capacitor connected to a 250  resistor and charged up until there are 9 V across the capacitor
a) What is the time constant for the discharging of the capacitor?
 = RC = 900 mF x 250  = 225 s
b) What is the voltage across the resistor at the moment the capacitor is connected and starts
discharging? How much current is flowing through the resistor at that moment?
VR = VC = 9 V
V
9V
I=R=
= 0.036 A
250 
c) What is the voltage across the resistor after the capacitor has been discharging for one time
constant? How much current is flowing through the resistor at that moment?
t =  = 225 s
-t
-225s
(
(
RC)
225 s)
VR = VC = V0e
= (9 V) e
= 3.31 V
V 3.31 V
I=R=
= 0.013 A
250 
d) What is the voltage across the resistor after the capacitor has been discharging for two time
constants? How much current is flowing through the resistor at that moment?
t = 2 = 450 s
-t
-450s
(
(
RC)
225 s)
VR = VC = V0e
= (9 V) e
= 1.22 V
V 1.22 V
I=R=
= 0.0049 A
250 
e) What is the voltage across the resistor after the capacitor has been discharging for three time
constants? How much current is flowing through the resistor at that moment?
t = 3 = 675 s
-t
-675s
(
(
RC)
225 s)
VR = VC = V0e
= (9 V) e
= 0.45 V
V 0.45 V
I=R=
= 0.0018 A
250 
14. A 30  flash bulb on a disposable camera is powered by a 0.25 F capacitor, which is charged by
connecting the circuit to a 6.5 V battery.
a) The flash bulb won’t fire until the capacitor is charged to at least 6 V. How long does the capacitor
have to charge to reach that level?
-t


RC

VC = V01 – e
-t


(0.25 F)(30 )

6 V = 6.5 V1 – e
-t


(0.25 F)(30 )

0.923 = 1 – e
( )
(
(
)
)
-t
(
(0.25 F)(30 ))
–0.077 = –e
-t
 ((0.25 F)(30

))

ln(0.077) = lne
-2.565 =
-t
(0.25 F)(30 )
t = 19.23 s
b) When you take a picture and the flash bulb is set off, it throws a switch that disconnects the battery
and connect the two ends of the capacitor directly. If the flash bulb is set off just when it is charged
enough to reach minimum voltage, how much current will flow through the resistor (flash bulb) as the
capacitor starts to discharge?
VR = VC = 6 V
V 6V
I=R=
= 0.2 A
30 
c) How much power is used in the flash bulb resistor as the capacitor starts to discharge?
P = IV = 0.2 A x 6 V = 1.2 W
d) If the flash only lasts for 0.12 s, how much energy is used by the flash? (Assume the current and
voltage remain essentially constant.)
E = P·t = 1.2 W · 0.12 s = 0.144 J