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]]>The theoretical results are obtained from the formula of Ohm’s law: V = IR. The experimental verification is provided for a metal film 1 kΩ (±0.05%). We have used a high-quality resistor with negligible tolerance value so as to reduce the tolerance error.

You might like: 5 Error Sources in Ohm’s Law Experiment [How to avoid them]

Voltage | Theory current | Experimental current |
---|---|---|

1 V | 1 mA | 1.000 mA |

2 V | 2 mA | 1.997 mA |

3 V | 3 mA | 3.000 mA |

4 V | 4 mA | 4.000 mA |

5 V | 5 mA | 5.001 mA |

6 V | 6 mA | 6.003 mA |

7 V | 7 mA | 7.000 mA |

8 V | 8 mA | 7.999 mA |

9 V | 9 mA | 9.839 mA |

10 V | 10 mA | 10.000 mA |

The little problem in our calculations arises due to improper handling of multimeter probes. You can learn the complete method to perform the Ohm’s experiment here and can calculate the current values by using the Ohm’s law calculator.

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]]>In fact, you can never match the theoretical calculations with practical values.

However, you can take some precautions to closely match the values.

Today’ you’ll learn the 5 error sources which are responsible for misleading readings. You’ll learn to keep you and your equipment safe by avoiding the blunders. You’ll also learn to obtain quite accurate readings. Let’s start off by understanding the types of errors.

Scientific measurement and instrumentation errors are often classified into three types:

- Personal errors: Mistakes made by the user due to his inexperience.
- Systematic: The faults in the instrument itself and the faults which may occur due to environmental conditions.
- Random errors: An accidental error whose cause is unknown. (We’ll ignore it here).

Generally, a personal error is an outright mistake which is made by the person himself. For example, you ignore a digit while taking observations. In case of Ohm’s law, you can commit a personal error by:

The ammeter is used to measure the current. It always connects in series with the circuit. Wrong connecting the ammeter will damage the instrument.

The voltmeter measures the potential difference between two points. It connects in parallel to the circuit. Wrong connecting the voltmeter will yield wrong readings.

Wrong measurements usually happen due to careless handling behavior. Carefully take the readings to avoid the errors.

Carbon and metal film resistors are the most popular class of resistors which are employed in our labs. Such resistors have a tolerance value which ranges between 0.05-20%. The leftmost band of carbon resistors indicates the possible tolerance of resistance. A silver band indicates a tolerance of 10%, the golden band indicates 5% and brown band indicates 1%. More tolerance means your resistance, and thus the voltage/current will fluctuate away from the theoretical value.

You have two choices to bypass this error.

* Use a brown [1%] or grey [0.05%] band resistor *which has low tolerance value and thus will provide a lower error.

* Measure the resistance first* and base your theoretical formula calculations on this value.

Your multimeter is the actual tool which measures the electrical quantities. While low-quality multimeters yield wrong observations, they are equally dangerous. Again you have two choices.

A variable power supply displays the output voltages on its main screen. For the time being, the accuracy of components decreases and your supply might display wrong results. Such cases are common in general labs where supplies are used thousands of times.

Use your multimeter to confirm the actual volts coming out of power supply.

Let’s summarize our results:

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]]>Two or more resistors are simply added in series to obtain the equivalent resistance.

The general formula is:

*R _{eq} = R_{1} + R_{2 }+ R_{3}+ … + R_{n}*

For two resistors:

**R _{eq} = R_{1} + R_{2}**

In such case, the resistor R in formulas of Ohm’s law will be replaced by * R_{eq}*.

Let’s do a simple example for learning it:

A 12-volt battery power two series resistors of 5-ohm and 30-ohm. Find the current flowing through the series circuit.

So a current of 0.342 A will flow through it.

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]]>The post Ohm’s Law Triangle appeared first on Ohm Law.

]]>The above triangle can easily be used to obtain three equations:

Voltage = [Current] * [Resistance]

Current = [Voltage] / [Resistance]

Resistance = [Voltage] / [Current]

or visually;

Let’s use it in some examples:

A 22 kΩ resistor has 10 mA current flowing through it. Use the Ohm’s triangle to find the value of connected voltage source.

From the left triangle, we obtain V = IR which states that voltage is directly proportional to resistance and current.

Here V = 10 mA * 22 kΩ = 220 V. So value of connected source is 220 V.

Let’s use the second * Ohm’s law triangle* in another example.

The protection relay at a power system requires 26 mA current for triggering purposes. Using ohmmeter the resistance of relays is found to be 200 ohms. Find the status of the relay when the source voltage is 5.2 V.

Solution: From the statement, I = V/R = 5.2 V / 200 Ω = 0.026 A or 26 mA which perfectly matches the operating value and hence relay will operate.

Let’s do another example:

The fuel meter of a car uses the gauge requires 10 mA for the proper operation. A 12 V battery powers the gauge meter of the car. Recommend an appropriate resistor which should be employed in the meter for proper operations.

Here I = 10 mA, V = 12 V. The * third triangle from above table* can be used to size the resistor. Here R = 12 V / 10 mA = 1.2 kΩ

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]]>The post Ohm’s Law Quiz MCQs with Answers appeared first on Ohm Law.

]]>1. The statement which correctly represents Ohm’s law:

- V = IR
- V = R/I
- R = VI
- I = R/V
- Correct answer: 1. V = IR

2. A 10 ohms resistor is powered by a 5-V battery. The current flowing through the source is:

- 10 A
- 50 A
- 2 A
- 0.5 A
- Correct answer: 4. 2 A
- Solution: From I = V/R = 5-V/10 ohm = 0.5 A

3. If V = 50 V and I = 5 A, then R = ___:

- 50 Ω
- 5 Ω
- 10 Ω
- 2 Ω
- Correct answer: 3. 10 Ω
- Solution: From R = V/I = 50 V/ 5A = 10 Ω

4. If P = 50 watt and R = 2 ohms, then I = ___?

- 50 A
- 5 A
- 10 A
- 2 A
- Correct answer: 3. 5 A
- Solution: From I = √(P/R) = √(50 V/ 2A) = √25 A = 5 A

5. Unit of voltage is:

- Volt
- Watt
- Coulomb
- Ampere
- Correct answer: 1. Volt

6. Unit of current is:

- Volt
- Watt
- Coulomb
- Ampere
- Correct answer: 4. Ampere

7. Unit of power is:

- Volt
- Watt
- Coulomb
- Ampere
- Correct answer: 2. Watt

8. Unit of resistance is:

- Volt
- Watt
- Ohms
- Ampere
- Correct answer: 3. Ohms

9. If V = 10 V and R = 15 kΩ, then I = ___?

- 0.666 mA
- 666 µA
- 0.66 A
- a & b
- Correct answer: 4. a & b
- Solution: Here I = V/R = 10 V / 15 kΩ = 0.666 mA = 666 µA

10. If I = 5 A and R = 10 Ω, then P = ___?

- 50 watts
- 250 watts
- 350 watts
- 500 watts
- Correct answer: 2. 250 watt
- Solution: Here P = I2R = (5 A)2 * 10 Ω = 250 watts

11. Components which obey Ohm’s law are known as:

- Resistors
- Ohmic components
- Non-ohmic components
- None of these
- Correct answer: 2. Ohmic components

12. Ohmic components have a __________ V-I curve:

- Straight line VI curve
- Parabolic VI curve
- Non-linear VI curve
- Sinusoidal VI curve
- Correct answer: 1. Straight line VI curve

13. An electric kettle with 50-Ω heat element is powered by a 230-V wall outlet. The current flowing through kettle is:

- 0.217 A
- 4.6 A
- 10.86 A
- 11500 A
- Correct answer: 2. 4.6 A

14. The water heating rod draws 10 A current when connected to certain power source. The resistance of rod is 12 Ω. The source voltage is:

- 1.2 V
- 120 V
- 1200 V
- None of these
- Correct answer: 2. 120 V

15. A relay with 100 Ω resistance requires 50 mA to for operation. When connected to a 4 V source the relay will:

- Operate
- Not operate
- Correct answer: 2. Not operate
- Solution: At 4 V, the current will I = V/R = 40 mA, which is not enough for the operation

16.The potential difference across a 5 kΩ is 12 V. Find the current flowing through the resistor:

- 60 mA
- 2.4 mA
- 1.77 mA
- 0.998 mA
- Correct answer: 2. 2.4 mA
- Solution: I = V/R = 12 V / 5 kΩ = 2.4 mA

17. An electrical bulb draws 5 A current when connected to a 100-V wall outlet, the resistance of bulb is:

- 5 Ω
- 20 Ω
- 100 Ω
- 500 Ω
- Correct answer: 2. 20 Ω

18. An unknown resistor has 5 mA current flowing through it when 10 volts are applied to it. For the same resistor, the application of 20 volts will result in the current flow of:

- 2 mA
- 5 mA
- 10 mA
- 50 mA
- Correct answer: 3. 10 mA
- Solution: Here R = 10 volts / 5 mA = 2 kΩ, Now for 20 volts, I = 20 volts/ 2 kΩ = 10 mA

Prefixes are numbers which represent the mathematical quantity in the power of ten. We already shared basic of prefixes in Ohm’s law article. In previous sections two prefixes ‘m’ and ‘k’ are introduced. This section of MCQs involves other prefixes as well. Let’s start learning:

19. If I_{1} = 50 mA, R_{1} = 50 kΩ, then V_{1} = __:

- 1 kV
- 2.5 kV
- 5 kV
- 10 kV
- Correct answer: 2. 2.5 kV
- Hint: 1k = 1000, 1m = 0.001

20. If I_{2} = 100 µA, R_{2} = 50 kΩ, then V_{2} = __:

- 5000 V
- 500 V
- 5 V
- 0.5 V
- Correct answer: 3. 5 V
- Hint: 1µ = 0.000 001

21. If I_{3} = 100 µA, R_{3} = 50 MΩ, then V_{3} = __:

- 5000 V
- 500 V
- 5 V
- 0.5 V
- Correct answer: 1. 5000 V
- Hint: 1M = 1 000 000

22. If V_{4} = 50 V, R_{4} = 10 GΩ, then I_{4} = __:

- 5 A
- 5 mA
- 5 µA
- 5 nA
- Correct answer: 4. 5 nA
- Hint: 1G = 1 000 000 000, 1 n = 0. 000 000 001

23. If V_{5} = 99 kV, R_{5} = 33 GΩ, then I_{5} = __:

- 5 A
- 5 mA
- 5 µA
- 5 nA
- Correct answer: 4. 5 nA

24. If V_{6} = 22 V, R_{6} = 11 kΩ, then I_{6} = __:

- 5 A
- 5 mA
- 5 µA
- 5 nA
- Correct answer: 4. 5 nA
- Hint: 1G = 1 000 000 000, 1 n = 0. 000 000 001

25. If V_{7} = 10 V, I_{7} = 1 μA, then R_{7} = __:

- 10 kΩ
- 10 MΩ
- 0.1 kΩ
- 0.1 MΩ
- Correct answer: 4. 10 MΩ

27. If V_{8} = 15 kV, I_{8} = 2 nA, then R_{7} = __:

- 7500 kΩ
- 7500 GΩ
- 30 kΩ
- 30 GΩ
- Correct answer: 4. 7500 GΩ

28. If V_{9} = 50 V, I_{9} = 25 μA, then R_{9} = __:

- 2 MΩ
- 2 GΩ
- 1250 MΩ
- 1250 GΩ
- Correct answer: 4. 2 GΩ

Color coding technique is used to represent the values of resistors in the form of colored bands. 4, 5 or 6 bands are designed over the bodies of resistors. While 4 band resistors are most commonly available we’ll use them in next problems. The first and second bands represent a number while the third and fourth one represents a multiplier.

29. A resistor is color-coded with four bands, the first one being brown, second, black, third red, and fourth gold. The resistor connects to a 10 V source. Find the current flowing through the resistor.

- 1 mA
- 10 mA
- 100 mA
- 1A
- Correct answer: 2. 10 mA
- Solution: The resistor can be color coded from the chart as shown below. It is 1k Ohms resistor. Neglecting the tolerance we can easily use the Ohm’s statement I = V/R to find the current.

30 If 1st band = Yellow; 2nd band = Violet; 3rd band = Brown; and V = 47 V, then I = ___?

- 0.1 A
- 0.1 mA
- 100 mA
- 470 mA
- Correct answer: 1. 0.1 A
- Solution: From color coding of resistors the resistor is 470 ohms. Now I = V/R = 47/470 = 0.1 A

31. If 1st band = Red; 2nd band = Red; 3rd band = Red; and V = 44 V, then I = ___?

- 20 A
- 20 mA
- 44 A
- 44 mA
- Correct answer: 2. 20 mA
- Solution: From color coding of resistors the resistor is 2.2 kohms. Now I = V/R = 44/2.2 k = 20 mA

32. If 1st band = Brown; 2nd band = Black; 3rd band = Green; and V = 50 V, then I = ___?

- 1 mA
- 100 μA
- 5 mA
- 50 μA
- Correct answer: 4. 50 μA
- Here resistance = 1 MΩ

33. If 1st band = Brown; 2nd band = Black; 3rd band = Brown; and I = 50 mA, then by using * Ohm law* find the value of source voltage:

- 0.5 V
- 5 V
- 50 V
- 500 V
- Correct answer: 2. 5 V
- Here resistance = 100 ohms

34. If 1st band = Brown; 2nd band = Black; 3rd band = Orange; and I = 50 mA, then by using the formula of *Ohm’s law* find the value of voltage source which powers the source:

- 0.5 V
- 5 V
- 50 V
- 500 V
- Correct answer: 4. 500 V
- Here resistance = 10k ohms

35. If 1st band = Orange; 2nd band = Orange; 3rd band = Orange; and I = 10 mA, then by using the formula of *Ohm’s law* find the value of voltage source which powers the source:

- 3 V
- 33 V
- 330 V
- 500 V
- Correct answer: 3. 330 V
- Here resistance = 33k ohms

This portion of quiz involves series circuits. A series circuit contains two or more in which head of the one joins the tail of other and there is no other connection in between them. To solve such circuits simply add the resistors and obtain a single equivalent resistor, then apply the law to find either current or voltage.

36. Two resistors having values 5 ohms and 10 ohms are connected in series to a 10 V source. Find the current flowing through the circuit:

- 0.66 A
- 1.5 A
- 2 A
- 2.5 A
- Correct answer: 1.5 A
- Solution: R
_{total}= R_{1}+ R_{2}= 5 ohms + 10 ohms = 15 ohms, Now I = V/R_{total}= 10 V / 15 ohms = 0.66 A

37. Two resistors having values 18 ohms and 50 ohms are connected in series to an unknown source. An ammeter is connected to the circuit which reads 2 A. Find the input voltage to the circuit:

- 36 V
- 100 V
- 136 V
- 168 V
- Correct answer: 136 V
- Solution: R
_{total}= R_{1}+ R_{2}= 18 ohms + 50 ohms = 68 ohms, Now V = I*R_{total}= 2A * 68 ohms = 136 V

A parallel circuit involves the configuration in which two resistors have their heads joined at common point and tail at another common point. The solution of two or more resistors in parallel follows a simple formula:

R_{total} = 1/((1/R_{1}) + (1/R_{2}))

38. Two parallel resistors both having their values 28 ohms are connected in parallel. The overall current provided by the 28 V source is.

- 1 A
- 2 A
- 4 A
- 8 A
- Correct answer: 2. 2 A
- Solution: R
_{total}= 1/((1/28 Ω) + (1/28 Ω)) = 14 ohms, I = V/R = 28/14 = 2 A

39. Two parallel resistors both having their values 50 and 60 ohms are connected in parallel. The overall current provided by the 100 V source is:

- 1.72 A
- 2.88 A
- 3.66 A
- 4.52 A
- Correct answer: 3. 66 A
- Solution: R
_{total}= 1/((1/50 Ω) + (1/60 Ω)) = 27.27 ohms, I = V/R = 100/27.27 = 3.66 A

40. Repeat the above problem for resistor values of 30 ohms each when the source is 60 V.

- 0.25 A
- 0.5 A
- 2 A
- 4 A
- Correct answer: 4. 4 A

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]]>V ∝ I

Where ‘V’ is voltage and ‘I’ is current and ‘R’ is constant of proportionality.

R is actually resistance of conductor which is assumed constant.

Removing this proportionality we get:

V = IR

We can use this formula to find current and resistance as well:

I = V/R

and R = V/I

You can easily memorize the above equations from the triangle.

- Draw a triangle.
- Divide it into 3 parts
- Mention V on top side
- Mention I on bottom left side
- Mention R on the bottom right side.
- To find V place your thumb on V and observe I and R. Here we get equation IR or better I.R
- To find I place your thumb on I and observe V and R. Here V is in numerator and R is in denominator we get equation I = V/R
- Finally, place your thumb on R and observe V and I. We get R = V/R

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]]>The post 5 Practical Applications of Ohm’s Law in Daily Life appeared first on Ohm Law.

]]>The electrical heater is a commonly used appliance in winters. Provided with the resistance of heater coil and applied voltage, We can calculate the power supplied to this heater. Let’s assume that resistance of heater coil is 5 ohm and input voltages are 120V. We can use the formula from Ohmic Wheel: P = V^{2}/R to find the power, P = 120^{2}/5 ohm = 2880 watt. This power can then be multiplied with time to calculate the electricity bill at our premises.

The speed control of conventional fans is achieved by using a potentiometer. A potentiometer is a variable resistance. A circular knob on the component can be rotated to achieve a variable resistance on the output terminals. For any specific value of input, we can calculate the resistance, current and thus power flowing through Ohm’s Law.

Most electronic devices in our daily life such as Mobile phone chargers, Laptop chargers, Internal circuits of devices and DC Power supplies all require a specific amount of current. Using Ohm’s Law we can calculate the amount of resistance that would be connected in that circuit for achieving any specific value of current.

Fuses and circuit breakers are the protection components which connect in series with the electronic devices. Fuses/CB’s are usually rated in Amperes. The selection of fuse for any specific rating of the component is calculated by Using Ohm’s Law. For example, you have an electric kettle which has an internal element of 10 ohms. The applied voltages are 100V. Now using Ohm’s law the current comes out to be I = V/R = 100V/10 ohm = 10A. Now while designing Fuse you are required to keep the rating at 10A.

Same as above.

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]]>The post Ohm’s Law Lab Report [With Graph, Observations and Verification] appeared first on Ohm Law.

]]>To verify that voltage and current are directly proportional using a 1kΩ resistor.

- Variable DC Power supply
- 1kΩ resistor (Color code Brown, Black, Red, Gold)
- Breadboard
- Connecting wires (Jumper wires)
- Ammeter

Setup the circuit diagram as shown below:

- Implement the circuit on the breadboard as shown in circuit diagram.
- Initially set the VDC supply to 0 V and note the current on ammeter.
- Increase the voltage to 1 V, observe the ammeter and note the readings.
- Repeat the above step for 2 V, 3V … 10 V.
- Plot a graph for measured readings.
- Compare the plot with theoretical calculations.

Make two columns (the third one for Sr. no is optional).

The theoretical calculations are achieved from the original formula, V = IR. You can also verify them from the basic Ohm’s law calculator.

The real (practically calculated – red color) vs the theoretical values (blue) are plotted on the graph. While most theoretical values are identical to practical values, one might note a difference of red dot for 2.99 mA and 6.99 mA.

We learned that current and voltage hold a direct relationship for resistive components. (They are linearly proportional).

Question: What is Variable DC supply?

Answer: A variable DC (direct current) supply provides variable output voltages. It contains a rotary knob which can be rotated to achieve our desired output voltages. Its symbol is similar to dc battery with the difference of an arrowhead over it.

Question: What is ammeter?

Answer: It is the current measuring instrument which is used in series to the circuit. It measures current and displays the amperes on LCD.

Question: What is the resistor, and what is resistor color coding?

Answer: Resistor is an electrical component which is used to control the flow of electrical current. Carbon resistors are the most commonly available type of resistors. They have color bands over their bodies which are used to identify the amount of resistance they possess.

Question: What are jumper wires?

Answer: They are special types of wires which can be easily mounted and used on the breadboards.

Question: Is Ohm’s Law important to learn?

Answer: Yes Ohm’s law is a fundamental law of Electrical Engineering and it’s very important. You can easily learn each and everything about the law from our website.

Question: Why practical results are different from theoretical results?

Answer: While no scientific-practical can perfectly match the theory, the actual results depend on environmental conditions, the accuracy of equipment under test. The quality of resistor also impacts the measurements, while in our lab report we got excellent results, you might get different results based on your theory.

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]]>The post Ohm’s Law Statement appeared first on Ohm Law.

]]>Mathematical statement of Ohm’s Law: **V = IR**

The above two lines explain the original statement of Ohm’s law. We can expand the law to derive the relationship between voltage, current, and resistance.

**For a constant resistance, the amount of current flowing through a circuit is directly proportional to its voltage.**

Now, V ∝ R

Let’s consider two cases for a better illustration.

Example # 1: A potential difference of 5 volts is applied to a 10 Ω resistor. Find the current flowing through it.

Solution: From Ohm’s formula, I = V/R = 5 V / 10 Ω = 0.5 A

Example # 2: A potential difference of 10 volts is applied to 10 Ω resistor (Since resistor is constant). Find the current passing through it.

Solution: I = V/R = 10 V / 10 Ω = 1 A

By comparing above cases one can understand that for a constant resistance (10 Ω) the amount of current increases when we increase the voltage.

**The electrical current flowing through any circuit is inversely proportional to its resistance.**

Now, for a constant voltage, statement will become: I = V/R → I ∝ 1/R

The current is inversely proportional to the connected resistor. The two examples below better illustrate the above statement.

Example # 3: A voltage source of 5 V connects to a 10 Ω resistor, find the current flowing through it.

Solution: From formula, I = V/R = 5 V / 10 Ω = 0.5 A.

Example # 4: A voltage source of 5 V connects (Since the source is constant), with a 20 Ω resistance, find the current flowing through it.

Solution: I = 0.25 A.

From above two examples, one can understand that by increasing voltage the current decreases.

**The amount of voltage dropped across any circuit is directly proportional to its resistance.**

Now, V ∝ R

The two cases below better illustrate this,

Example #5: Find the amount of voltage applied across 10 kΩ resistor when a current of 5 mA flows through it.

Solution: V = IR = 5 mA * 10 kΩ = 50 V

Example # 6: Find the amount of voltage applied across the 20 kΩ resistor when a current of 5 mA (since current is constant) flows through it.

Solution: V = IR = 5 mA * 20 kΩ = 100 V.

By comparing both cases one can understand that amount of voltage increases when the resistor increases.

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]]>

Let’s take a look at Ohm’s wheel:

We can use the wheel to extract 12 formulas (3 for each voltage, current, power, and resistance):

- V = IR
- V = P/I
- V = (P.R)
^{1/2} - P = VI
- P = V
^{2}/R - P = I
^{2}R - I = P/V
- I = (P/R)
^{1/2} - I = V/R
- R = V/I
- R = P/I
^{2} - R = V
^{2}/P

The Ohm’s wheel is the visual representation of Ohm’s formulas. It can be used to calculate different parameters.

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]]>