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Voltage, Current and Resistance
Foundations in Engineering – WV Curriculum, 2002
Foundations in Engineering Content Standards and Objectives 2436.8.3 Explain the relationship between current, voltage, and resistance,
and use appropriate meters to measure each. Voltage, Current, and Resistance Over the last few days, you have learned about three different terms used in electricity and electronics: voltage, current, and resistance. Voltage is the pressure that pushes electrons through a conductor and is measured in volts (V). It results from the potential difference between a positive and a negative terminal. Voltage can be measured with a voltmeter and is always the measurement of the difference between two different parts of a circuit. The individual voltage measurements across all parts of a series circuit add up to the voltage across the power source. V1 = V2 + V3
The voltage measurements in a parallel circuit across each branch will be the same as the voltage across the power source.
V1 = V2 = V3
V1
V2
V3
V1
V2
V3
Current is the quantity of electrons that flow through a conductor. Current is measured in amperes (A). In microelectronics, you often deal with very small current values expressed as milliamperes (mA), which equals 1/1000th of an ampere. Current is affected by the voltage between two points, but it is also affected by the resistance from different materials. Resistance is the opposition to the flow of electrons through a conductor. The type of material affects its resistance, as well as the length of the conductor and its temperature. Resistance is measured in ohms (R). It is often necessary to reduce the current flow in a circuit to protect components. Voltage divider is the term used to describe the method used for reducing the supply voltage from the battery. Electricians use specially constructed components called resistors that withstand the heat they produce. Each resistor has a power rating consisting of a coded set of colored bands that tells the electrician the value of the resistor. (see Transparency TM 10-3-1)
To determine the value of a resistor, the numbers represented by the first two colored bands are multiplied by the number represented by the third band.
Orange = 3, blue = 6 and red = 100 36 x 100 = 3600 R
Resistors can be combined to produce a needed resistance. If they are combined in series, the total resistance equals the sum of all of the individual resistances.
RT = R1 + R2 + R3
50 + 100 + 1000 = 1150 R If they are combined in parallel, the total resistance is determined by the following formula: 1 1 1 1 RT = R1 + R2 + R3
1 1 1 20 10 1 31 1 50 + 100 + 1000 = 1000 + 1000 + 1000 = 1000 = 32.2 R
Often a circuit calls for different levels of voltage for different situations. A variable resistor, or potentiometer, is designed to adjust the flow of electricity
much like water is controlled through a tap. A 200 ohm variable resistor can be adjusted from 0 ohms to 200 ohms.
Ohm’s Law The units of volts, amperes, and ohms are mathematically related in Ohm’s Law: 1 volt of pressure through 1 ohm of resistance equals 1 ampere of current. This mathematical law is used to calculate an unknown value of voltage, current, or resistance when the other two values are known. There are three ways of stating the relationship, and the graphic below will help you to remember the formulas: V = Voltage in volts I = Current in amperes R = Resistance in ohms V = I x R I = V/R R = V/I Example: Find the current in an automobile light bulb that has a resistance of 16 ohms and operates on 12 volts. Choose the correct formula: Current (I) = Voltage (V)/Resistance (R) Substitute known values: I = 12/16 = 3/4 = .75 amps Activity 1. Have 10 Tronix kits, 10 multimeters, and 10 calculators for student group use
during this activity. 2. Discuss voltage and current and how to calculate voltage measurements
within a circuit. 3. Have students complete Handout 10-3-1. 4. Discuss resistance, how to determine the value of a resistor, (Transparency
TM 10-3-1) and how to calculate the resistance in both series and parallel circuits.
5. Have students complete Handout 10-3-2. 6. Demonstrate how to use a multimeter to measure voltage and current.
Remember that voltage is measured ACROSS the load being tested, and
V
R I
current is measured IN the circuit. 7. Hand out the Tronix kits and multimeters for groups of students to use. Have
students complete Handout 10-3-3, experimenting with resistors, variable resistors and multimeters. Review activity.
8. Discuss Ohm’s Law and how to determine an unknown value of voltage,
current or resistance when the other two values are known. (Transparency TM 10-3-2) A great website with a student tutorial and a lot of applets for students to do on-line experiments is at www.mste.uiuc.edu/users/noemi/ohm/menu.html
9. Have students complete Handout 10-3-4 and review in class.
Resistor Color Coding Transparency TM 10-3-1
.01 Sil 10% .1 Gold 5% 0 0 0 1 1 1 1 10 1% 100 2 2 2 100 2% 50 3 3 3 1K 25 4 4 4 10K 15 5 5 5 100K .5% 6 6 6 1M .25% 7 7 7 10M .1% 8 8 8 9 9 9
1ST D
IGIT
2ND D
IGIT
3RD D
IGIT
MU
LTIP
LIER
TOLE
RA
NC
E
TE
MP
CO
EF
4 BAND
6 BAND
5 BAND
25K 5%
276K 5%
460K 1%
Ohm’s Law Transparency TM 10-3-2
There are three ways of stating Ohm’s Law,
and the graphic below will help you to remember the formulas:
V = Voltage in volts
I = Current in amperes R = Resistance in ohms
V = I x R
I = V/R
R = V/I
V
R I
Figuring Voltage Handout 10-3-1
Calculate the voltage measurement in each of the following circuits. Make Sure to show your work. Remember: For Series circuits - the voltage is added VT = V1 + V2 + V3 For Parallel circuits - each branch remains the same VT = V1 = V2 = V3 1.
2.
3.
4.
36 V
V
V?
?
?
6 V
36V
h
Lamps = 3 V eac?
3 V
V
V
6V
3VV
24 V
3V
Figuring Voltage Handout 10-3-1 Answer Sheet
Calculate the voltage measurement in each of the following circuits. Make Sure to show your work. Remember: For Series circuits - the voltage is added VT = V1 + V2 + V3 For Parallel circuits - each branch remains the same VT = V1 = V2 = V3 1.
V1 = V2 + V3 + V4 V1 = 6 + 3 + 3 V1 = 12
2.
V1 = V2 + V3 + V4 + V5 36 = 3 + 3 + 3 + V5 36 = 9 + V5 V5 = 27
3. V1 = V2 = V3 36 = V2 36 = V3
4.
h
1
3
24 V
3 2 4 3 2 5
36 V
2 V6 V2
4
V1 = V2 + V3 + V4 24 = 2 + V3 + 6 24 = 8 + V3 V3 = 16 V3 = V5 16 = V5
3
5
36
Lamps = 3 V eac
V
V16
3
V1
V
V
V1
V4
V
V V V V VVV2
VV3
V
Working with Resistors Handout 10-3-2
1-4. Figure the value of each resistor below from their color code. Make sure to
show your work. 5. Are th 6. a. W b. If yohms to resistors
R G Oe currents in both circuits the same? Wh
hat is the total resistance of this circuit? ou wanted to be able to vary the resist
the answer in a. above, what would you h with?
10V
60 R
40 R10 V
30 R
20 R10 V
50R
W B R
Bl Br YP B G R
y or why not?
ance in this circuit from 0 ave to replace the current
100 R
7. Are the currents in both circuits the same? Why or why not? 8. What size resistor would you need to replace these
resistors in this circuit? 9. Figure the value of this series combination of resistors below from their color
code. Make sure to show your work.
10. Figure the value of this parallel combination of resistors below from their color
code. Make sure to show your work.
30 R
12 V
30 R
60 R
12 V
R G R G O R Y P B
RGR
G B R
Br R G Br
400 R
400 R
12 V
200 R
Challenge Yourself! Any network of resistors in a circuit can be replaced by a single resistor. Its value is called the network’s equivalent resistance. You can find the equivalent resistance of networks by simplifying the circuit step by step. See if you can figure out the equivalent resistance for the circuit below. It helps to draw a new circuit for each step. Show all work. Step 1. Simplify the 2 parallel 400 ohm resistors. Step 2. Simplify the series circuit. Step 3. Simplify the parallel circuit.
9V 400 500
300
400
Working with Resistors Handout 10-3-2 Answer Sheet
1-4. Figure the value of each resistor below from their color code. Make sure
to show your work. 25 x 1K = 25,000 705 x 100 = 70,500 or 25K 61 x 10K = 61,000 90 x 100 = 9,000 or 61K or 9K
5. Are the currents in both circuits the same? Why or why not? Yes, because in a series circuit, the values of the resistors is added together
40 + 60 = 100 6. a. What is the total resistance of this circuit? 100 ohms
b. If you wanted to be able to vary the resistance in this circuit from 0 ohms to the answer in a. above, what would you have to replace the current resistors with? 100 ohm variable resistor (potentiometer)
10V
100 R 60 R
40 R10 V
30 R
20 R 10 V
50R
7. Are the currents in both circuits the same? Why or why not?
No, because in a parallel circuit, the inverse of the values of the resistors are added together.
3/60 = 1/20 R should equal 20
8. What size resistor would you need to replace these resistors in this circuit? 100 ohms 1/400 + 1/400 + 1/200 = 4/400 = 1/100
8. Figure the value of this series combination of resistors below from their color code. Make sure to show your work.
(25 x 100)+(47 x 1)+(53 x 100) = 2500 + 47 + 5300 = 7847 R 10. Figure the value of this parallel combination of resistors below from their color
code. Make sure to show your work.
1/R = 1/2500 + 1/ 5000 + 1/1250 = 2/5000 + 1/5000 + 4/5000 1/R = 7/5000 = 1/714.2 R = 714.2
R G R G O R Y P B
RGR
GBR
Br R G Br
30 R
12 V
30 R
60 R
12 V
400 R
400 R
12 V
200 R
Challenge Yourself!
Any network of resistors in a circuit can be replaced by a single resistor. Its value is called the network’s equivalent resistance. You can find the equivalent resistance of networks by simplifying the circuit step by step. See if you can figure out the equivalent resistance for the circuit below. It helps to draw a new circuit for each step. Show all work. Step 1. The 400 R and 400 R resistors are in parallel. They can be replaced by a single 200 R resistor. 1/R = 1/400 + 1/400 = 2/400 = 1/200 R = 200 Step 2. The 300 R and 200 R resistors are in series. They can be replaced by a single 500 R resistor. R = 300 + 200 = 500 Step 3. You are now left with a circuit with two 500 R resistors in parallel. Again, parallel 1/RT = 1/R1 + 1/R2 1/R = 1/500 + 1/500 = 1/R = 2/500 1/R = 1/250 R = 250
9 200 500
300
9 500 500
9
9V 400 500
300
400
250
Variable Resistors Handout 10-3-3
Materials and Equipment 2 6-volt bulbs 1 6-volt battery 1 0 to 10 volt voltmeter 1 1K variable resistor 1 2K variable resistor 1 1K resistor connecting wires baseboard Procedures 1. Assemble the circuit shown, using 2 6-volt bulbs. Measure the voltage across
Bulb A using a voltmeter set at 10V or above. Then measure the voltage across Bulb B. What is happening in the circuit?
Voltage A = __________ Voltage B = __________ 2. Replace Bulb A with a 200-ohm variable resistor. Adjust the variable
resistor and record how dim the bulb goes, and how bright it goes. 3. Replace the 200-ohm variable resistor with a 100-ohm variable resistor.
What difference does it make compared to the task above? 4. Adjust the variable resistor until the bulbs are just glowing. Now measure the
voltage across Bulb B and Variable Resistor A. What did you find out? 5. Now replace Bulb B with a 1K resistor. Record the voltage range across 1K
resistor B when Variable Resistor A is at either end of its movement by adjusting the control knob. Can you come up with a statement about the relationship of voltage to resistance?
A B
Ohm’s Law Handout 10-3-4
Ohm’s Law V = I * R
1. What current would flow through the resistor in the circuit if the voltage across it were 5V? 2. What voltage would be needed to make 5 mA flow through the resistor? (Hint:
Ohm’s Law expresses current in Amps.) 3. To what value would you have to set the resistor at to make the current equal
to 10 mA, if the voltage were 8 V?
4. The lamp is designed to work at a current of 60 mA. Its resistance is 100 ohms. What voltage would you have to put across the lamp to make it work at its working current? How much voltage would be needed if another lamp was added in series?
500
5. An LED uses only 2V and can take only 10mA without being damaged. Find
the value of the resistor that will be needed in a series circuit if a 6-volt battery is used.
6. Find the value of the resistor that will be needed in a series circuit if two 9-volt
batteries are used. Express value in kilo-ohms.
Critical Thinking 7. Now think about the relationship between the current, voltage and resistance.
Show what would happen to the voltage if the current was doubled. What would the resistance be if the current was doubled? What would the current be if the voltage were doubled? (Hint: substitute these figures into your equations: voltage = 6, current = 3 and resistance = 2).
6V
8. Below is a simplified schematic of the heater blower motor in a car. The
switch can connect the circuit to each of the three pathways. The low pathway contains all three resistors in series. The medium pathway contains just two of the resistors. The high pathway only passes through one resistor. Use what you know about Ohm’s law to explain why the heater blower motor runs at a different power on each of these
settings. 9. Household voltage varies around the world. In the US we have a standard
voltage of 110 volts. Some other countries don’t even have a standard voltage. What would happen if you plugged a 110-volt hair dryer from the US into a 240-volt outlet in England? Explain your answer using Ohm’s Law.
Ohm’s Law Handout 10-3-4 Answer Sheet
Ohm’s Law V = I * R
1. What current would flow through the resistor in the circuit if the voltage across it were 5V? Looking for current (I), so formula is I = V/R I = 5/500 = 1/100 A = .01A = 10 mA (multiple by 1000)
2. What voltage would be needed to make 5 mA flow through the resistor? (Hint: Ohm’s Law expresses current in Amps.) Looking for voltage (V), so formula is V = I * R Change 5 mA to A by dividing by 1000 = .005 A V = .005 A * 500 = 2.500 V
3. To what value would you have to set the resistor at to make the current equal to 10 mA, if the voltage were 8 V? Looking for resistance (R), so formula is R = V/I Change 10 mA to A by dividing by 1000 = .010 A R = 8/.010 = 800R
4. The lamp is designed to work at a current of 60 mA. Its resistance is 100 ohms. What voltage would you have to put across the lamp to make it work at its working current? How much voltage would be needed if another lamp was added in series? Looking for voltage (V), so formula is V = I * R Change 60 mA to A by dividing by 1000 = .06 A V = .06 A * 100 = 6 V
a. Change 10 mA to A 10/1000 = .01 A b. R = V/I R = 4/.01 = 400 ohms
500
5. An LED uses only 2V and can take only 10mA without being damaged. Find the value of the resistor that will be needed in a series circuit if a 6-volt battery is used.
c. Find voltage across resistor. VT = V1+ V2 6 = V1 + 2 Voltage across resistor = 4 volts
a. Find voltage across resistor. VT = V1+ V2 18 = V1 + 2 Voltage across resistor = 16 volts
b. Change 10 mA to A 10/1000 = .01 A c. R = V/I R = 16/.01 = 1600 ohms or 16K ohms
6. Find the value of the resistor that will be needed in a series circuit if two 9-volt batteries are used. Express value in kilo-ohms.
7. Critical Thinking
Now think about the relationship between the current, voltage and resistance. Show what would happen to the voltage if the current was doubled. What would the resistance be if the current was doubled? What would the current be if the voltage were doubled? (Hint: substitute these figures into your equations: voltage = 6, current = 3 and resistance = 2).
Original Double Current What happens? V = I * R 6 = 3 * 2 V = 6 * 2 = 12 current doubled voltage doubles
R = V/I 2 = 6/3 R = 6/6 = 1 current doubled resistance halved I = V/R 3 = 6/2 I = 12/2 = 6 voltage doubled current doubled
6V
8. Below is a simplified schematic of the heater blower motor in a car. The
switch can connect the circuit to each of the three pathways. The low pathway contains all three resistors in series. The medium pathway contains just two of the resistors. The high pathway only passes through one resistor. Use what you know about Ohm’s law to explain why the heater blower motor runs at a different power on each of these
settings.
As the resistance is cut down, the current increases. Therefore, because the “low” path has the most resistance, it will run the slowest. As the resistance is made lower, the current will be higher, and the blower will run faster.
9. Household voltage varies around the world. In the US we have a standard
voltage of 110 volts. Some other countries don’t even have a standard voltage. What would happen if you plugged a 110 volt hair dryer from the US into a 240 volt outlet in England? Explain your answer using Ohm’s Law.
As the voltage increases, so does the current. Too much current through the hair dryer would burn up the components.
