Ohm’s Law Sample Problems
1.
In a circuit consisting of a voltage source and a resistor, describe what happens to the
current when: (remember the % difference or % change equation – it is very useful)
=
% change
a.
new value − reference value
× 100
reference value
the voltage tripled:
Three-times more voltage (pressure) will produce three-times more current or
200% increase in current flow.
V
5V
15V
I = ⇒
=1A ; (three-times voltage)
=3A
R
5Ω
5Ω
% change
=
b.
3 A − 1A
× 100
= 200%
1A
the voltage is reduced by 75%:
Current is decreased by 75% - only 25% of original value
5V reduced 75% = 1.25V
(75% of 5V = 0.75 x 5V = 3.75V and 5V – 3.75 = 1.25V)
(Reduced means subtract that percentage from the original value)
1.25V
V
0.25 A
I =⇒
=
5Ω
R
0.25 A 1
3
= meaning that
or 75% has been lost .
1A
4
4
or using the % change formula to determine the percentage of change:
0.25 A − 1A
% change =
× 100 =
−75%
1A
c.
the resistance is doubled:
Twice the resistance (opposition to current flow) means the current will be half
its original value or 50% of its value.
5V
5V
V
I = ⇒
=1A ; (two-times resistance)
=0.5 A
5Ω
10Ω
R
0.5 A − 1A
% change =
× 100 =
−50%
1A
d.
the resistance reduced by 35%:
Remember that current vs. voltage is linear and that current vs resistance is
inversely related (not linear but exponentially). Reduce resistance by 35% will
increase current by 54%
35% reduction means the resistor will maintain 65% of its original value:
R= 0.65 × 10Ω= 6.5Ω
V
10V
I =⇒
1A ;
=
R
10Ω
=
% change
e.
1.54 A − 1A
=
× 100 54%
1A
the voltage is doubled and the resistance is cut in half:
Current will be four times larger or it will quadruple or increase by 300%.
10V
V
1A ;
I =⇒
=
10Ω
R
=
% change
15.
10V
1.54 A
=
6.5Ω
20V
4A
=
5Ω
4 A − 1A
× 100
= 300%
1A
Determine the current in each case:
a.
V = 5V, R = 1.0Ω
V
5V
I = ⇒
=5 A
R
1Ω
b.
V = 15V, R = 10Ω
15V
V
1.5 A
I =⇒
=
10Ω
R
c.
V = 50V, R = 100Ω
V
50V
I = ⇒
=0.5 A ⇒ 500 × 10 −3 A ⇒ 500mA
R
100Ω
d.
V = 30V, R = 15kΩ
I =
e.
V
30V
⇒
= 0.002 A ⇒ 2 × 10 −3 A ⇒ 2mA
R
15k Ω
V = 250V, R = 5.6MΩ
V
250V
I =⇒
0.000044642 A ⇒ 44.64 × 10 −6 A ⇒ 44.64 µ A
=
R
5.6M Ω
19.
A resistor is connected across the terminals of a 25V source. Determine the current in
the resistor if the color code is yellow, violet, orange, silver.
First, determine the value of the resistor: 47,000Ω ±10% or 47k Ω ±10%. This means
that this resistors value can be anywhere between 10% below and above its color-coded
value. Subtract 10% from the CCV and add 10% to the CCV to see the range of values.
10% of 47, 000
=
Ω 4, 700Ω
47, 000Ω − 4700
=
Ω 42, 300Ω or 42.3k Ω
47, 000Ω + 4700
=
Ω 51, 700Ω or 51.7k Ω
Because the resistor was stated to have this 10% tolerance, the circuit must be
examined using the low and high possibilities for the 47kΩ resistor. We will be
determining the maximum and minimum current values that could be in the circuit.
Es
25V
V
R
R
I(max) =
⇒
R1
42.3kohm
25V
591.02 µ A
=
42.3k Ω
I(max) calculator operation
23.
V
R
R
I(min) =
⇒
25V
483.56 µ A
=
51.7k Ω
I(min) calculator operation
Calculate the voltage for each value of I and R:
using Ohm’s Law relationship of V = I x R:
a.
I = 2A,
R = 18Ω
36V
b.
I = 5A,
R = 56Ω
280V
c.
I = 2.5A, R = 680Ω
1.7kV
d.
I= 0.6A, R = 47Ω
28.2V
e.
I = 0.1A, R = 560Ω
56V
26.
Assign a voltage value to each source in the following circuits, to obtain the indicated
amounts of current in each.
a.
VR = I × R ⇒ 3mA × 27k Ω ⇒ 3 × 10 −3 A × 27 × 103 Ω = 81V
b.
VR = I × R ⇒ 5 µ A × 100M Ω ⇒ 5 × 10 −6 A × 100 × 106 Ω = 500V
c.
VR = I × R ⇒ 2.5 A × 47Ω = 117.5V
Homework Assignment #4
problems: 31,33, and two bonus
problems
31.
A filament of a lamp has a certain amount of resistance, represented by an equivalent
resistance shown in the following diagram. If the lamp operates with 120V and 0.8A of
current, what is the resistance of its filament when it is on?
33.
By varying the rheostat ( variable resistor) in a circuit you can change the amount of
current. The setting of the rheostat is such that the current is 750mA. (a) What is the
resistance value of this setting? (b) To adjust the current to 1A, to what resistance
value must you set the rheostat? (c) Can there be a problem with this circuit setup?
(a). Use Ohm's equation to determine resistance (being this is the only load in the
circuit it has to use all the source voltage).
(b).
At 1A of current the resistance will be:
(c). The rheostat (variable resistor) is the only load in this circuit and if the resistor is
adjusted so that it has no resistance there will not be a device that can use the applied
energy from the 100V source in this circuit. The current will not be controlled and
damage would result in the source, connection wires, and/or the rheostat.
Bonus.
A variable voltage source is connected to a 100Ω resistor. Starting at 0V and
increasing the voltage in 10V steps up to 100V, determine the current at each voltage.
Plot a graph of V vs. I. Is the graph a straight line? What does the graph indicate?
Graphs are used to convey information is a graphical way, so that lists of numbers do
not have to interpreted. The numeric values are all contained in the graph or chart. Start
by using Ohm's Law to calculate the expected current for each voltage step: I = V/R.
The graph indicates a straight line and shows Voltage Vs. Current.
Applied
Voltage
0V
10V
20V
30V
40V
50V
60V
70V
80V
90V
100V
Circuit
Current
0mA
100mA
200mA
300mA
400mA
500mA
600mA
700mA
800mA
900mA
1A
Ohm’s Law Bonus Problems
31.
A filament of a lamp has a certain amount of resistance, represented by an equivalent resistance
shown in the following diagram. If the lamp operates with 120V and 0.8A of current, what is the
resistance of its filament when it is on?
33.
By varying the rheostat ( variable resistor) in a circuit you can change the amount of current.
The setting of the rheostat is such that the current is 750mA. (a) What is the resistance value of
this setting? (b) To adjust the current to 1A, to what resistance value must you set the
rheostat? (c) Can there be a problem with this circuit setup?
(a) Use Ohm's equation to determine resistance (being this is the only load in the circuit
it has to use all the source voltage).
(b) At 1A of current the resistance will be:
(c) The rheostat (variable resistor) is the only load in this circuit and if the resistor is
adjusted so that it has no resistance there will not be a device that can use the
applied energy from the 100V source in this circuit. The current will not be controlled
and damage would result in the source, connection wires, and/or the rheostat.
Bonus: A variable voltage source is connected to a 100Ω resistor. Starting at 0V and increasing the
voltage in 10V steps up to 100V, determine the current at each voltage. Plot a graph of V vs. I.
Is the graph a straight line? What does the graph indicate?
Graphs are used to convey information is a graphical way, so that lists of numbers do not have to
interpreted. The numeric values are all contained in the graph or chart. Start by using Ohm's
Law to calculate the expected current for each voltage step: I = V/R. The graph indicates a
straight line and shows Voltage Vs. Current.
Applied
Voltage
0V
10V
20V
30V
40V
50V
60V
70V
80V
90V
100V
Circuit
Current
0mA
100mA
200mA
300mA
400mA
500mA
600mA
700mA
800mA
900mA
1A