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SS1: PHYSICS – 3RD TERM

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  1. Production of Electric Current | Week 1
    6 Topics
    |
    1 Quiz
  2. Electric Current | Week 2
    5 Topics
    |
    1 Quiz
  3. Electrical Resistance of a Conductor | Week 3
    5 Topics
    |
    1 Quiz
  4. Particulate Nature of Matter | Week 4
    5 Topics
    |
    1 Quiz
  5. Crystalline and Non-crystalline Substances | Week 5
    3 Topics
    |
    1 Quiz
  6. Elastic Properties of Solids | Week 6 & 7
    4 Topics
    |
    1 Quiz
  7. Fluids at Rest & in Motion | Week 8 & 9
    6 Topics
    |
    1 Quiz
  8. Solar Collector
    3 Topics
    |
    1 Quiz



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A German scientist Georg Simon Ohm investigated the relationship between voltage or potential difference and current flowing along a metallic conductor. His finding is known as Ohm’s law which is stated as follows;

The current flowing along a metallic conductor at constant temperature is directly proportional to the potential difference across its end provided other physical conditions remain constant.

If V represents the potential difference in volts, I represent the current flowing in amperes, then

\( \scriptsize V \propto I \)

V = IR

R = \( \frac {V}{I} = \frac {Potential\; difference}{current}\)

R is a constant of proportionality known as resistance of a conductor. It depends on the nature of the conductor

∴ V = IR

I = \(\frac{V}{R}\)

The unit of resistance is Ohm (Ω).

Ohm is defined as the resistance of a conductor when a potential difference or a voltage of one volt is applied across the conductor causes a current of one ampere to flow through the conductor.

Limitations of Ohm’s Law: Ohm’s law holds for metals and certain materials, but some do not obey Ohm’s law. These materials include diodes, transistors, rectifiers, semiconductors, rare gas, and acids.

Let’s apply ohms law in the following examples;

Example 1

Calculate the current produced when a 350-volt cell is connected across a 12 ohms resistor.

Solution:

Values in the question:

Voltage V = 350 volt,  Resistance R = 5 ohms, current I = ?

Formula:                      

Voltage V = IR

Substitute the values given into the equation             

350 = I x 5

Make current the subject of the formula:          

current I = \( \frac{350}{5}\)  .    

Current I = 70 Amperes

Now go ahead and solve this on your own.

Exercise 1
Determine the value of resistance that will cause a voltage of 125 Volt to generate 2.5 Amperes of Current.

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Exercise 1

Determine the value of resistance that will cause a voltage of 125 Volt to generate 2.5 Amperes of Current.

Values in the question:

Voltage V = 1250 V,  Resistance R = ?, current I = 2.5 A

Formula:

Voltage V = IR

Substitute the values given into the equation   

125 = 2.5 x R

Make R the subject of the formula:  

R = \( \frac{125}{2.5}\) 

R = 50Ω

error: