Chapter 35
Worksheet: Circuits and Ohm’s Law
EQUATIONS
ELECTRICAL CIRCUIT SYMBOLS
1. Draw a circuit schematic (diagram) to include a
50.0 V battery, an ammeter, and a resistance of
10.0 Ω in series.
A
+
50.0 V
-
10.0 Ω
a. What is the reading on the ammeter?
I 
V
I 
R
50 V
10 . 0 
 5A
A
+
50.0 V
-
10.0 Ω
b. In which direction is the current flowing?
Conventional current (+)
A
+
50.0 V
10.0 Ω
-
Electron flow (-)
2. How much current flows through a radio speaker
that has a resistance of 4.0 Ω when 16 V is
impressed across the speaker?
I 
V
R
I 
16 V
4 .0 
 4 .0 A
Draw a circuit diagram of the circuit described in the question
above. Include a 6 V battery, an ammeter (labeled with value
of current), and a resistance of 3.0 Ω (the speaker). Also
label the direction of the conventional (+) current.
Conventional current
2A
A
+
6V
3.0 Ω
-
I 
V
R
I 
6V
3
I  2A
4. The following questions
pertain to the circuit
diagramed to the right.
a. Lamp A reads a voltage
of 12 V. What is the
voltage of lamp B?
4. The following questions
pertain to the circuit
diagramed to the right.
a. Lamp A reads a voltage
of 12 V. What is the
voltage of lamp B?
Since this is a parallel circuit, the voltage in each
branch of the circuit is equal to the voltage source.
V s  V1  V 2  V 3 ......
Therefore, Voltage is equal to 12 volts in each lamp
b. If the ammeters on both
branches read the same
amount, what does this tell
you about the resistance of
the two branches?
b. If the ammeters on both
branches read the same
amount, what does this tell
you about the resistance of
the two branches?
Since this is parallel circuit, the resistance in each
branch can be calculated with the following
equation:
V
V
IA 
RA 
Rearrange and get
RA
IA
Since the voltage and current are the same in each
branch, the resistance will be the same
c. If the current flowing in
the first branch was 4.0 A
and 6.0 A in the second
branch, what would the
total current in the circuit
be?
c. If the current flowing in
the first branch was 4.0 A
and 6.0 A in the second
branch, what would the
total current in the circuit
be?
Total current in a parallel circuit can be calculated
with the following equation:
I  I A  I B  I C ......
I  4 .0 A  6 .0 A
I  10 A
5. Draw a series circuit diagram showing a 6.0 V
battery, a resistor, & an ammeter reading of 2.0 A.
2.0 A
A
6.0 V
a. Label: the size of the resistor, the direction of
conventional current, the (+) and (-) terminals of the
battery.
2.0 A
A
6.0 V
a. Label: the size of the resistor, the direction of
conventional current, the (+) and (-) terminals of the
battery.
Conventional current
2.0 A
A
+
6.0 V
3Ω
-
I 
Vs
R
R 
Vs
I
R 
6 . 0V
2 .0 A
 3
b. Add a voltmeter to you diagram and indicate the
potential difference across the resistor.
2.0 A
A
6.0 V
V s  V1  V 2  V 3 .....
V
6.0 V
Therefore Vs = V1 = 6.0 V
6. Draw a circuit diagram showing a heater with a
resistance of 6 Ω, and a potential difference
source of 24.0 V.
A
24.0 V
6Ω
V
a. Calculate the current through the resistance
I 
Vs
R
I 
24 . 0V
I  4 .0 A
6
A
24.0 V
6Ω
V
b. What thermal energy is supplied by the
heater in 10 seconds? (HINT- use the equation
E = I2Rt to determine energy)
A
24.0 V
6Ω
E  I Rt
2
E   4 . 0 A  6 . 0  10 s 
2
V
E  960 J
7. Use the circuit diagram to the
right to answer the following
questions.
a. What is the current flowing
through this series circuit if the
total resistance is 20 Ω?
I 
Vs
R
I 
24V
20 
I  1 .2 A
b. What would the voltage across
each of the three bulbs be? What
could you say about the
brightness of each of the bulbs?
V s  V1  V 2  V 3 .....
24 V  8V  8V  8V
Therefore each bulb would have 8 volts and the
same brightness
c. If two of the bulbs had a total
resistance of 15 Ω, what would
the resistance of the third bulb
be?
I 
Vs
R
R  20 
1 .2 A 
24 . 0V
R
R 
24 . 0V
1 .2 A
If the total resistance equals 20
ohms, then the remaining resistor
would have a value of 5 ohms
(20 – 15 = 5)
d. What would be the current
flowing through the circuit be is
the voltage source was 6.0 V, and
each of the lamps had a
resistance of 2 Ω?
2Ω
6.0V
2Ω
2Ω
First calculate total
resistance
R  R A  R B  RC
R  2  2  2  6
Then plug into equation
for current
I 
Vs
R
I 
6V
6
 1A
8. Draw a circuit diagram showing three 10 Ω
resistors connected in parallel and placed across a
60.0 V battery.
A
60.0 V
10Ω
10Ω
10Ω
a. What is the equivalent resistance of the
parallel circuit?
1
R

1
RA

1
RB

1
1
RC
R

1
10

1

10
1
1
10
R

3
10
A
60.0 V
10Ω
Cross multiply and get
10Ω
3 R  10
10Ω
R  3 .3 
b. What is the current through the entire circuit?
R= 3.3 Ω (from
earlier problem)
I 
V
I 
R
60 . 0V
3 .3
 18 A
A
60.0 V
10Ω
10Ω
10Ω
c. What is the current through each branch of
the circuit?
IA 
V
RA
IA 
60 . 0V
10 
 6 .0 A
A
60.0 V
10Ω
10Ω
10Ω
Since the voltage and resistance are the same in
each branch they would have the same current
9. Draw a circuit diagram showing the following: a
800.0 Ω resistor, a 40 Ω resistor, and a 20 Ω resistor
connected in parallel and connected across a 24.0
V battery.
A
24.0 V
800Ω
40Ω
20Ω
a. What is the equivalent resistance of the
parallel circuit?
1
R

1
RA

1
RB

1
1
RC
R

1
800

1

40
1
1
20
R

61
800
A
24.0 V
800Ω
40Ω
Cross multiply and get 61 R  800
20Ω
R  13 
b. What is the current through the entire circuit?
From above
problem
R  13 
I 
V
I 
R
24 . 0V
13 
 1 .9 A
A
24.0 V
800Ω
40Ω
20Ω
c. What is the current through each branch of
the circuit?
24 . 0V
V
IA 
 0 . 03 A
IA 
800 
RA
A
24.0 V
IB 
800Ω
24 . 0V
40 
 0 .6 A
40Ω
IC 
20Ω
24 . 0V
20 
 1 .2 A
10. Answer the following
questions about the circuit
to the right.
a. What do each of the
4 voltmeters read?
6V
In a parallel circuit, the voltage is the same through
each branch and divided by the number of resistors
in each branch. Therefore:
VA = 6 V
VB = 6 V
VC = 3 V and VD = 3 V (voltage is split between
the two resistors in that branch
b. If each of the resistors
are identical, and the total
current flowing through this
parallel circuit is 12.0 A,
what is the total resistance
of this circuit?
I 
V
R
R 
V
I
6V
R 
6 . 0V
12 . 0 A
 0 .5 