cC
Chapter 23
a What is the current through each resistor?
1. Three 12.0-fl resistors are connected in
series to a 50.0-V power source
a. What is the equivalent resistance of the
circuit?
=
45.0 V
is.oa
=
=12.00+12.00+12.00
= 36.0
h. What is the current in the circuit?
y
I?
= 50.0 V
36.04)
=1.39A
a What is the voltage drop across each
resistor?
=
=3.OOA
3. Two resistors are connected in series to a
power source The voltage drop across the
first resistor is 5.40 V and the voltage drop
across the second resistor is 9.80 V. The cur
rent through the circuit is 1.20 A.
a. What is the resistance of each of the
resistors?
v
7
R54J)v=
=
=4.501)
V
v=rn
1
4
eov
(1.39 AX12.0 Ii)
=16.7V
=
.4
=
2. Three 15.0-fl resistors are connected in
parallel to a 45.0-V power source
a What is the equivalent resistance of the
circuit?
1
is.oii +
3
=
1
+
1
15.04)
1
=
1
R = 5.00 a
b. What is the current in the circuit?
b. What is the equivalent resistance of the
circuit?
+R
1
R=R
2
=4.500+8.17(1
=12.70
4. What is the equivalent resistance of the
circuit shown below? What is the current
in the circuit? What is the voltage drop
across the two resistors wired in parallel?
flnd the equivalent resistance of the
resistors In parallel.
1
i-I
MO!
l.20A
=8.170
R
=
5.40 V
1.20 A
1
1
1
1
45.OV
s.ooti
=9.OOA
Physics: Principles and Problems
Supplemental Problems Answer KEy
179
wer Key
Chapter 23 continued
Using the equivalent resistance of the
resistors in parallel, find the resistance
of the resistors in series.
Req2
=
6.
A 12-fl and an 18-fl resistor are connected
to a 48-V power source.
a. What is the equivalent resistance of the
circuit if the resistors are connected in
series?
1 + Reqi
R
=88 ft + 6.0 ft
R= R
1 +
=94 ft
=
12 ft +18 ft
=
1 ft
3.OX1O
The current in the circuit:
V
Req2
—
I=
b. What is the equivalent resistance of the
circuit if the resistors are connected in
parallel?
I
Req2
1_i + 1
RP
9
F?2
120 V
94 a
1
—
=1.3A
—
R= 7.2 ft
7. A voltage divider is made from a 9.0-V bat
V= IRq
=
7.8 V
1
18.0 ft
36
—
(1.3 A)(6.0 ft)
+
5
The voltage drop across the two
resistors:
V
Reqi
i
=
12.0 ft
tery. Two resistors are connected in series to
the battery. if one resistor has a resistance of
24 ft and the voltage drop across the other
resistor must be 4.0 V. what is the resistance
of the second resistor?
88 ci
Find the current in the circuit.
Let R be the unknown resistance.
12
120V
ci
Then (9.0 V)
12 fl
=
-I-=
&OV
=
+ V
24
=
(4.0 V) + 1(24 II)
0.21 A
I
5. A 10.0-fl resistor, a 20.0-fl resistor, and a
30.0-fl resistor are wired in parallel and
connected to a 15.0-V power source. What
is the equivalent resistance of the circuit?
1
R
R
1
—
—
2
R
io.o a
+
3
R
Find the resistance of the second
resistor.
4.0 V
0.21 A
=19ft
1
1
+
20.0 ft
30.0 ft
11
60
R= 5.45 ft
180
Supplemental Problems Answer Key
Physics: Principles arid Problems
AwerKey
.
Chapter 23 continued
8. A circuit is constructed, as shown in the
figure below. The voltmeter reads 63.0 V.
10.
a. Which resistor dissipates the most
energy per second?
A I0.0-( resistor and a 20.0—12 resistor ai e
connected in series with a potent iom eicr
and a 9.0-V battery.
a. What should the potentiometer he set
at kr a total equivalent lesistance of
50.0 12 in this circuit?
R=V
I
V
1=
RT
R
10.0 f
b.
1.8 A
=
P= IR
=
+
20.0
20.0
=
If the potentiometer is set at .32 0 12,
what would he thc cumient in this cmruit?
20.0
C.
V = IR
1 + R
I(R
2 + R
)
3
=
(1.8 A)(42 II + 36 f + 54 )
=
240 V
=
62.0 II
=
9OV
=014A
62.0 II
If’ the potelili mdcc “eie turned so that
the resistance increases, what wi mid
happen to the curt em
Since the resistors are connected
in series, as R increases, RT wiN
increase by the same amount. From
Ohm’s law, we see that as resistance
and current are inversely propor
tional so as RT increases, the cur
rent in the circuit will decrease.
+ R
3
=
± 32.0
Vsource
RT
1=
b. What is the voltage of the power source?
=
—
+R=10.0.+
2
RT=Rl +R
(1.8 A)
R
2
Thus, the resistor with the highest
resistance will dissipate the most
energy per second. So, the 54-.
resistor dissipates the most energy
per second.
R
1 + R
R
2 +
=50.0k-2
Rp=RT—Rl —R
63.0 V
36
—
=
11.
A piecc’ of lab equipment must In’ Connec
ed to a standard 6.1) V div ccli, I tic maim
ftr the equipment says ii at his dcvi c has
an internal resistance of 0 10 12 amid aimnof
handle more than 2.5 \ of mum ‘cut.
a. What value oi m’csms!or can von (1111 ccl
in series with this (le\ mdc that would
allow it to he con m1c ted
9.
‘l’hree identical resistors are connected in
parallel across a power source. ‘Iheir equiva
lent resistance is 8.00 12. What is the resis
tance of each resistor?
Let R
1
Then,
=
2
R
=
=
-
3
R
+
-
=
+
R
to
itn’ po’. r
SOUrce?
From Ohm’s law,
RT
=
Vsource
—
—
RT
R
=
6.OV
2.5 A
—
2.412
—
R + Rinternai
=
RT
=
2.3 12
Rinternai
=
2.4 !!
010 12
-
3
R
R= 24.0 f
Physics: Principles and Problems
So pplenieutai Pii+lei,i’
tl%iiL’1 kei
181